Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 13 - Mean And Variance Of A Random Variable

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Page No 31.14:

Question 1:

Which of the following distributions of probabilities of a random variable X are the probability distributions?
(i)

X :	3	2	1	0	−1
P (X) :	0.3	0.2	0.4	0.1	0.05

(ii)

X :	0	1	2
P (X) :	0.6	0.4	0.2

(iii)

X :	0	1	2	3	4
P (X) :	0.1	0.5	0.2	0.1	0.1

(iv)

X :	0	1	2	3
P (X) :	0.3	0.2	0.4	0.1

Answer:

(i) P (X = 3) + P (X = 2) + P (X = 1) + P (X = 0) + P (X = $-$ 1)
    = 0.3 + 0.2 + 0.4 + 0.1 + 0.05
    =1.05 > 1
It is not the probability distribution of random variable X.

(ii) P (X = 0) + P (X = 1) + P (X = 2)
     = 0.6 + 0.4 + 0.2
     = 1.2 > 1
It is not the probability distribution of random variable X.

(iii) P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
      = 0.1 + 0.5 + 0.2 + 0.1 + 0.1
      = 1
It is the probability distribution of random variable X.

(iv) P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
      = 0.3 + 0.2 + 0.4 + 0.1
      = 1
It is the probability distribution of random variable X.

Page No 31.14:

Question 2:

A random variable X has the following probability distribution:

Values of X :	−2	−1	0	1	2	3
P (X) :	0.1	k	0.2	2k	0.3	k

Find the value of k.

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = $-$ 2) + P (X = $-$ 1) + P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

$\Rightarrow 0.1 + k + 0.2 + 2 k + 0.3 + k = 1 \Rightarrow 4 k + 0.6 = 1 \Rightarrow k = \frac{0.4}{4} = 0.1$

Page No 31.14:

Question 3:

A random variable X has the following probability distribution:

Values of X :	0	1	2	3	4	5	6	7	8
P (X) :	a	3a	5a	7a	9a	11a	13a	15a	17a

Determine:
(i) The value of a
(ii) P (X < 3), P (X ≥ 3), P (0 < X < 5).

Answer:

(i) Since the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

$\Rightarrow a + 3 a + 5 a + 7 a + 9 a + 11 a + 13 a + 15 a + 17 a = 1 \Rightarrow 81 a = 1 \Rightarrow a = \frac{1}{81}$

(ii) P (X < 3)

$= P (X = 0) + P (X = 1) + P (X = 2) = \frac{1}{81} + \frac{3}{81} + \frac{5}{81} = \frac{9}{81} = \frac{1}{9}$

P (X ≥ 3)

$P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = \frac{7}{81} + \frac{9}{81} + \frac{11}{81} + \frac{13}{81} + \frac{15}{81} + \frac{17}{81} = \frac{72}{81} = \frac{8}{9}$

P (0 < X < 5)
$P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = \frac{3}{81} + \frac{5}{81} + \frac{7}{81} + \frac{9}{81} = \frac{24}{81} = \frac{8}{27}$

Page No 31.14:

Question 4:

The probability distribution function of a random variable X is given by

x_i :	0	1	2
p_i :	3c³	4c − 10c²	5c − 1

where c > 0
Find: (i) c (ii) P (X < 2) (iii) P (1 < X ≤ 2)

Answer:

(i) We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) = 1

$\Rightarrow 3 c^{3} + 4 c - 10 c^{2} + 5 c - 1 = 1 \Rightarrow 3 c^{3} - 10 c^{2} + 9 c - 2 = 0 \Rightarrow (c - 1) (3 c^{2} - 7 c + 2) = 0 \Rightarrow (c - 1) (3 c - 1) (c - 2) = 0 \Rightarrow c = \frac{1}{3}, 1, 2 (Neglecting 1 and 2 as individual probability should not be greater than one)$

(ii) P (X < 2)

$= P (X = 0) + P (X = 1) = 3 c^{3} + 4 c - 10 c^{2} = \frac{1}{9} + \frac{4}{3} - \frac{10}{9} = \frac{1 + 12 - 10}{9} = \frac{3}{9} = \frac{1}{3}$

(iii) P (1 < X ≤ 2)

$= P (X = 2) = 5 c - 1 = \frac{5}{3} - 1 = \frac{5 - 3}{3} = \frac{2}{3}$

Page No 31.14:

Question 5:

Let X be a random variable which assumes values x₁, x₂, x₃, x₄ such that 2P (X = x₁) = 3P (X = x₂) = P (X = x₃) = 5 P (X = x₄). Find the probability distribution of X.

Answer:

Let P (X = x₃) = k. Then,
P (X = x₁) = $\frac{k}{2}$

P (X = x₂) = $\frac{k}{3}$

P (X = x₄) = $\frac{k}{5}$

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = x₁) + P (X = x₂) + P (X = x₃) + P (X = x₄) = 1

$\Rightarrow \frac{k}{2} + \frac{k}{3} + k + \frac{k}{5} = 1 \Rightarrow \frac{15 k + 10 k + 30 k + 6 k}{30} = 1 \Rightarrow \frac{61 k}{30} = 1 \Rightarrow k = \frac{30}{61}$

Now,

x_i	p_i
x₁	$\frac{k}{2}$ = $\frac{15}{61}$
x₂	$\frac{k}{3}$ = $\frac{10}{61}$
x₃	k = $\frac{30}{61}$
x₄	$\frac{k}{5}$ = $\frac{6}{61}$

Page No 31.14:

Question 6:

A random variable X takes the values 0, 1, 2 and 3 such that:
P (X = 0) = P (X > 0) = P (X < 0); P (X = −3) = P (X = −2) = P (X = −1); P (X = 1) = P (X = 2) = P (X = 3). Obtain the probability distribution of X.

Answer:

Let P (X = 0) = k. Then,
P (X = 0) = P (X > 0) = P (X < 0)
$\Rightarrow$ P (X > 0) = k
P (X < 0) = k

∴ P (X = 0) + P (X > 0) + P (X < 0) = 1

$\Rightarrow k + k + k = 1 \Rightarrow k = \frac{1}{3}$

Now,
P (X < 0) = k

$\Rightarrow P (X = - 1) + P (X = - 2) + P (X = - 3) = k \Rightarrow 3 P (X = - 1) = k [∵ P (X = - 1) = P (X = - 2) = P (X = - 3)] \Rightarrow P (X = - 1) = \frac{k}{3} \Rightarrow P (X = - 1) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} ∴ P (X = - 1) = P (X = - 2) = P (X = - 3) = \frac{1}{9} Similarly, P (X > 0) = k \Rightarrow P (X = 1) = P (X = 2) = P (X = 3) = \frac{1}{9}$

Thus, the probability distribution is given by

X_i	P_i
$-$ 3	$\frac{1}{9}$
$-$ 2	$\frac{1}{9}$
$-$ 1	$\frac{1}{9}$
1	$\frac{1}{9}$
2	$\frac{1}{9}$
3	$\frac{1}{9}$

Page No 31.14:

Question 7:

Two cards are drawn from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer:

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
$P (X = 0) = P (no ace) = \frac{^{48} C_{2}}{^{52} C_{2}} = \frac{2256}{2652} = \frac{188}{221} P (X = 1) = P (1 ace) = \frac{^{4} C_{1} \times^{48} C_{1}}{^{52} C_{2}} = \frac{192}{1326} = \frac{32}{221} P (X = 2) = P (2 aces) = \frac{^{4} C_{2}}{^{52} C_{2}} = \frac{6}{1326} = \frac{1}{221}$

Thus, the probability distribution of X is given by

X	P (X)
0	$\frac{188}{221}$
1	$\frac{32}{221}$
2	$\frac{1}{221}$

Page No 31.14:

Question 8:

Find the probability distribution of the number of heads, when three coins are tossed.

Answer:

Let X denote the number of heads in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
$P (X = 0) = P (T T T) = \frac{1}{8}, P (X = 1) = P (H T T or T T H or T H T) = \frac{3}{8} P (X = 2) = P (H T H or T H H or H H T) = \frac{3}{8}, P (X = 3) = P (H H H) = \frac{1}{8}$

Thus, the probability distribution of X is given by

X	P (X)
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$

Page No 31.14:

Question 9:

Four cards are drawn simultaneously from a well shuffled pack of 52 playing cards. Find the probability distribution of the number of aces.

Answer:

Let X denote the number of aces in a sample of 4 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take values 0, 1, 2, 3 and 4.
Now,
$P (X = 0) = P (no ace) = \frac{^{48} C_{4}}{^{52} C_{4}} P (X = 1) = P (1 ace) = \frac{^{4} C_{1} \times^{48} C_{3}}{^{52} C_{4}} P (X = 2) = P (2 aces) = \frac{^{4} C_{2} \times^{48} C_{2}}{^{52} C_{4}} P (X = 3) = P (3 aces) = \frac{^{4} C_{3} \times^{48} C_{1}}{^{52} C_{4}} P (X = 4) = P (4 aces) = \frac{^{4} C_{4}}{^{52} C_{4}}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{^{48} C_{4}}{^{52} C_{4}}$
1	$\frac{^{4} C_{1} \times^{48} C_{3}}{^{52} C_{4}}$
2	$\frac{^{4} C_{2} \times^{48} C_{2}}{^{52} C_{4}}$
3	$\frac{^{4} C_{3} \times^{48} C_{1}}{^{52} C_{4}}$
4	$\frac{^{4} C_{4}}{^{52} C_{4}}$

Page No 31.14:

Question 10:

A bag contains 4 red and 6 black balls. Three balls are drawn at random. Find the probability distribution of the number of red balls.

Answer:

Let X denote the number of red balls in a sample of 3 balls drawn from a bag containing 4 red and 6 black balls. Then, X can take the values 0, 1, 2 and 3.
Now,
$P (X = 0) = P (no red ball) = \frac{^{6} C_{3}}{^{10} C_{3}} = \frac{20}{120} = \frac{1}{6} P (X = 1) = P (1 red ball) = \frac{^{4} C_{1} \times^{6} C_{2}}{^{10} C_{3}} = \frac{60}{120} = \frac{1}{2} P (X = 2) = P (2 red balls) = \frac{^{4} C_{2} \times^{6} C_{1}}{^{10} C_{3}} = \frac{36}{120} = \frac{3}{10} P (X = 3) = P (3 red balls) = \frac{^{4} C_{3}}{^{10} C_{3}} = \frac{4}{120} = \frac{1}{30}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{6}$
1	$\frac{1}{2}$
2	$\frac{3}{10}$
3	$\frac{1}{30}$

Page No 31.14:

Question 11:

Five defective mangoes are accidently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.

Answer:

Let X denote the number of defective mangoes in a sample of 4 mangoes drawn from a bag containing 5 defective mangoes and 15 good mangoes. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
$P (X = 0) = P (no defective mango) = \frac{^{15} C_{4}}{^{20} C_{4}} = \frac{1365}{4845} = \frac{91}{323} P (X = 1) = P (1 defective mango) = \frac{^{5} C_{1} \times^{15} C_{3}}{^{20} C_{4}} = \frac{2275}{4845} = \frac{455}{969} P (X = 2) = P (2 defective mangoes) = \frac{^{5} C_{2} \times^{15} C_{2}}{^{20} C_{4}} = \frac{1050}{4845} = \frac{70}{323} P (X = 2) = P (3 defective mangoes) = \frac{^{5} C_{3} \times^{15} C_{1}}{^{20} C_{4}} = \frac{150}{4845} = \frac{10}{323} P (X = 3) = P (4 defective mangoes) = \frac{^{5} C_{4}}{^{20} C_{4}} = \frac{5}{4845} = \frac{1}{969}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{91}{323}$
1	$\frac{455}{969}$
2	$\frac{70}{323}$
3	$\frac{10}{323}$
4	$\frac{1}{969}$

Page No 31.15:

Question 12:

Two dice are thrown together and the number appearing on them noted. X denotes the sum of the two numbers. Assuming that all the 36 outcomes are equally likely, what is the probability distribution of X?

Answer:

Let X denote the sum of the numbers on two die. Then, X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Sample space : {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
                            (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
                            (3, 1), (3, 2), (3 ,3), (3, 4), (3, 5), (3, 6)
                            (4, 1), (4, 2), (4 ,3), (4, 4), (4, 5), (4, 6)
                          (5, 1), (5, 2), (5 ,3), (5, 4), (5, 5), (5, 6)
                          (6, 1), (6, 2), (6 ,3), (6, 4), (6, 5), (6, 6)}

Now,
$P (X = 2) = \frac{1}{36} P (X = 3) = \frac{2}{36} P (X = 4) = \frac{3}{36} P (X = 5) = \frac{4}{36} P (X = 6) = \frac{5}{36} P (X = 7) = \frac{6}{36} P (X = 8) = \frac{5}{36} P (X = 9) = \frac{4}{36} P (X = 10) = \frac{3}{36} P (X = 11) = \frac{2}{36} P (X = 12) = \frac{1}{36}$

Thus, the probability distribution of X is given by

X	P(X)
2	$\frac{1}{36}$
3	$\frac{2}{36}$
4	$\frac{3}{36}$
5	$\frac{4}{36}$
6	$\frac{5}{36}$
7	$\frac{6}{36}$
8	$\frac{5}{36}$
9	$\frac{4}{36}$
10	$\frac{3}{36}$
11	$\frac{2}{36}$
12	$\frac{1}{36}$

Page No 31.15:

Question 13:

A class has 15 students whose ages are 14, 17, 15, 14, 21, 19, 20, 16, 18, 17, 20, 17, 16, 19 and 20 years respectively. One student is selected in such a manner that each has the same chance of being selected and the age X of the selected student is recorded. What is the probability distribution of the random variable X?

Answer:

Here, X can take the values 14, 15, 16, 17, 19, 20 and 21.
Now,
$P (X = 14) = \frac{2}{15} P (X = 15) = \frac{1}{15} P (X = 16) = \frac{2}{15} P (X = 17) = \frac{3}{15} P (X = 18) = \frac{1}{15} P (X = 19) = \frac{2}{15} P (X = 20) = \frac{3}{15} P (X = 21) = \frac{1}{15}$

Thus, the probability distribution of X is given by

X	P(X)
14	$\frac{2}{15}$
15	$\frac{1}{15}$
16	$\frac{2}{15}$
17	$\frac{3}{15}$
18	$\frac{1}{15}$
19	$\frac{2}{15}$
20	$\frac{3}{15}$
21	$\frac{1}{15}$

Page No 31.15:

Question 14:

Five defective bolts are accidently mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.

Answer:

Let X denote the number of defective bolts in a sample of 4 bolts drawn from a bag containing 5 defective bolts and 20 good bolts. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
$P (X = 0) = P (no defective bolts) = \frac{^{20} C_{4}}{^{25} C_{4}} = \frac{4845}{12650} = \frac{969}{2530} P (X = 1) = P (1 defective bolt) = \frac{^{5} C_{1} \times^{20} C_{3}}{^{25} C_{4}} = \frac{5700}{12650} = \frac{114}{253} P (X = 2) = P (2 defective bolts) = \frac{^{5} C_{2} \times^{20} C_{2}}{^{25} C_{4}} = \frac{1900}{12650} = \frac{38}{253} P (X = 3) = P (3 defective bolts) = \frac{^{5} C_{3} \times^{20} C_{1}}{^{25} C_{4}} = \frac{200}{12650} = \frac{4}{253} P (X = 4) = P (4 defective bolts) = \frac{^{5} C_{4}}{^{25} C_{4}} = \frac{5}{12650} = \frac{1}{2530}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{969}{2530}$
1	$\frac{114}{253}$
2	$\frac{38}{253}$
3	$\frac{4}{253}$
4	$\frac{1}{2530}$

Page No 31.15:

Question 15:

Two cards are drawn successively with replacement from well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer:

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
$P (X = 0) = P (no ace) = \frac{48}{52} \times \frac{48}{52} = \frac{12 \times 12}{13 \times 13} = \frac{144}{169} P (X = 1) = P (1 ace) = \frac{4}{52} \times \frac{48}{52} = \frac{2 \times 12}{13 \times 13} = \frac{24}{169} P (X = 2) = P (2 aces) = \frac{4}{52} \times \frac{4}{52} = \frac{1 \times 1}{13 \times 13} = \frac{1}{169}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{144}{169}$
1	$\frac{24}{169}$
2	$\frac{1}{169}$

Page No 31.15:

Question 16:

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of kings.

Answer:

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
$P (X = 0) = P (no kings) = \frac{48}{52} \times \frac{48}{52} = \frac{12 \times 12}{13 \times 13} = \frac{144}{169} P (X = 1) = P (1 king) = \frac{4}{52} \times \frac{48}{52} = \frac{2 \times 12}{13 \times 13} = \frac{24}{169} P (X = 2) = P (2 kings) = \frac{4}{52} \times \frac{4}{52} = \frac{1 \times 1}{13 \times 13} = \frac{1}{169}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{144}{169}$
1	$\frac{24}{169}$
2	$\frac{1}{169}$

Page No 31.15:

Question 17:

Two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Answer:

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
$P (X = 0) = P (no ace) = \frac{48}{52} \times \frac{47}{51} = \frac{2256}{2652} = \frac{188}{221} P (X = 1) = P (1 ace) = \frac{4}{52} \times \frac{48}{51} + \frac{48}{52} \times \frac{4}{51} = \frac{384}{2652} = \frac{32}{221} P (X = 2) = P (2 aces) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{188}{221}$
1	$\frac{32}{221}$
2	$\frac{1}{221}$

Page No 31.15:

Question 18:

Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement, from a bag containing 4 white and 6 red balls.

Answer:

Let X denote the number of white balls in a sample of 3 balls drawn from a bag containing 4 white and 6 red balls. Then, X can take the values 0, 1, 2 and 3.
Now,
$P (X = 0) = P (no white ball) = \frac{^{6} C_{3}}{^{10} C_{3}} = \frac{20}{120} = \frac{1}{6} P (X = 1) = P (1 white ball) = \frac{^{4} C_{1} \times^{6} C_{2}}{^{10} C_{3}} = \frac{60}{120} = \frac{1}{2} P (X = 2) = P (2 white balls) = \frac{^{4} C_{2} \times^{6} C_{1}}{^{10} C_{3}} = \frac{36}{120} = \frac{3}{10} P (X = 3) = P (3 white balls) = \frac{^{4} C_{3}}{^{10} C_{3}} = \frac{4}{120} = \frac{1}{30}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{6}$
1	$\frac{1}{2}$
2	$\frac{3}{10}$
3	$\frac{1}{30}$

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Question 19:

Find the probability distribution of Y in two throws of two dice, where Y represents the number of times a total of 9 appears.

Answer:

It is given that Y denotes the number of times a total of 9 appears on throwing the pair of dice.

When the dice is thrown 2 times, the possibility of getting a total of 9 is possible only for the given combinations:

(3, 6) (4, 5) (5, 4) (6, 3)

So, the total number of outcomes is 36 and the total number of favourable outcomes is 4.

Probability of getting a total of 9 = $\frac{4}{36} = \frac{1}{9}$

Probability of not getting a total of 9 = $1 - \frac{1}{9} = \frac{8}{9}$

If Y takes the values 0, 1 and 2, then

$P (Y = 0) = \frac{8}{9} \times \frac{8}{9} = \frac{64}{81} P (Y = 1) = \frac{1}{9} \times \frac{8}{9} + \frac{8}{9} \times \frac{1}{9} = \frac{16}{81} P (Y = 2) = \frac{1}{9} \times \frac{1}{9} = \frac{1}{81}$

Thus, the probability distribution of X is given by

Y	P(Y)
0	$\frac{64}{81}$
1	$\frac{16}{81}$
2	$\frac{1}{81}$

Page No 31.15:

Question 20:

From a lot containing 25 items, 5 of which are defective, 4 are chosen at random. Let X be the number of defectives found. Obtain the probability distribution of X if the items are chosen without replacement.

Answer:

Let X denote the number of defective items in a sample of 4 items drawn from a bag containing 5 defective items and 20 good items. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
$P (X = 0) = P (no defective item) = \frac{^{20} C_{4}}{^{25} C_{4}} = \frac{4845}{12650} = \frac{969}{2530} P (X = 1) = P (1 defective item) = \frac{^{5} C_{1} \times^{20} C_{3}}{^{25} C_{4}} = \frac{5700}{12650} = \frac{114}{253} P (X = 2) = P (2 defective items) = \frac{^{5} C_{2} \times^{20} C_{2}}{^{25} C_{4}} = \frac{1900}{12650} = \frac{38}{253} P (X = 3) = P (3 defective items) = \frac{^{5} C_{3} \times^{20} C_{1}}{^{25} C_{4}} = \frac{200}{12650} = \frac{4}{253} P (X = 4) = P (4 defective items) = \frac{^{5} C_{4}}{^{25} C_{4}} = \frac{5}{12650} = \frac{1}{2530}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{969}{2530}$
1	$\frac{114}{253}$
2	$\frac{38}{253}$
3	$\frac{4}{253}$
4	$\frac{1}{2530}$

Page No 31.15:

Question 21:

Three cards are drawn successively with replacement from a well-shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Determine the probability distribution of X.

Answer:

Let X denote the number of hearts in a sample of 3 cards drawn from a well-shuffled deck of 52 cards. Then, X can take the values 0, 1, 2 and 3.
Now,
$P (X = 0) = P (no heart) = \frac{39}{52} \times \frac{39}{52} \times \frac{39}{52} = \frac{27}{64} P (X = 1) = P (1 heart) = (\frac{13}{52} \times \frac{39}{52} \times \frac{39}{52}) + (\frac{39}{52} \times \frac{13}{52} \times \frac{39}{52}) + (\frac{39}{52} \times \frac{39}{52} \times \frac{13}{52}) = \frac{27}{64} P (X = 2) = P (2 hearts) = (\frac{13}{52} \times \frac{13}{52} \times \frac{39}{52}) + (\frac{39}{52} \times \frac{13}{52} \times \frac{13}{52}) + (\frac{13}{52} \times \frac{39}{52} \times \frac{13}{52}) = \frac{9}{64} P (X = 3) = P (3 hearts) = \frac{13}{52} \times \frac{13}{52} \times \frac{13}{52} = \frac{1}{64}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{27}{64}$
1	$\frac{27}{64}$
2	$\frac{9}{64}$
3	$\frac{1}{64}$

Page No 31.15:

Question 22:

An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.

Answer:

Let X denote the number of blue balls in a sample of 3 balls drawn from a bag containing 4 red and 3 blue balls. Then, X can take values 0, 1, 2 and 3.
Now,
$P (X = 0) = P (no blue ball) = \frac{4}{7} \times \frac{4}{7} \times \frac{4}{7} = \frac{64}{343} P (X = 1) = P (1 blue ball) = (\frac{3}{7} \times \frac{4}{7} \times \frac{4}{7}) + (\frac{4}{7} \times \frac{3}{7} \times \frac{4}{7}) + (\frac{4}{7} \times \frac{4}{7} \times \frac{3}{7}) = \frac{144}{343} P (X = 2) = P (2 blue balls) = (\frac{3}{7} \times \frac{3}{7} \times \frac{4}{7}) + (\frac{4}{7} \times \frac{3}{7} \times \frac{3}{7}) + (\frac{3}{7} \times \frac{4}{7} \times \frac{3}{7}) = \frac{108}{343} P (X = 3) = P (3 blue balls) = \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} = \frac{27}{343}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{64}{343}$
1	$\frac{144}{343}$
2	$\frac{108}{343}$
3	$\frac{27}{343}$

Page No 31.15:

Question 23:

Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the probability distribution of the number of successes, when getting a spade is considered a success.

Answer:

Let X denote the number of spades in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

$P (X = 0) = P (no spade) = \frac{^{39} C_{2}}{^{52} C_{2}} = \frac{741}{1326} = \frac{19}{34} P (X = 1) = P (1 spade) = \frac{^{13} C_{1} \times^{39} C_{1}}{^{52} C_{2}} = \frac{507}{1326} = \frac{13}{34} P (X = 2) = P (2 spades) = \frac{^{13} C_{2}}{^{52} C_{2}} = \frac{78}{1326} = \frac{1}{17}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{19}{34}$
1	$\frac{13}{34}$
2	$\frac{1}{17}$

Page No 31.15:

Question 24:

A fair die is tossed twice. If the number appearing on the top is less than 3, it is a success. Find the probability distribution of number of successes.

Answer:

Let X denote the event of getting a number less than 3 (1 or 2) on throwing the die. Then, X can take the values 0, 1 and 2.
Now,
$P (X = 0) = \frac{16}{36} = \frac{4}{9} P (X = 1) = \frac{16}{36} = \frac{4}{9} P (X = 2) = \frac{4}{36} = \frac{1}{9}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{4}{9}$
1	$\frac{4}{9}$
2	$\frac{1}{9}$

Page No 31.15:

Question 25:

An urn contains 5 red and 2 black balls. Two balls are randomly selected. Let X represent the number of black balls. What are the possible values of X. Is X a random variable?

Answer:

The possible values of X are 0, 1 and 2, i.e. no black ball, 1 black ball and 2 black balls.
Yes, X is a random variable.
A random variable is a real valued function having domain as the sample space associated with a random experiment.

Page No 31.15:

Question 26:

Let X represent the difference between the number of heads and the number of tails when a coin is tossed 6 times. What are possible values of X?

Answer:

Given: X = Number of heads − Number of tails

Number of heads	Number of heads	Number of heads − Number of tails
0	6	−6
1	5	−4
2	4	−2
3	3	0
4	2	2
5	1	4
6	0	6

Therefore, the possible values of X are :
−6, −4, −2, 0, 2, 4, 6

Page No 31.15:

Question 27:

From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.

Answer:

Let X denote the number of defective bulbs in a sample of 2 bulbs drawn from a lot of 10 bulbs containing 3 defectives and 7 non-defectives.Then X can take values 0, 1, 2.
Now,
$P (X = 0) = P (no defective bulb) = \frac{^{7} C_{2}}{^{10} C_{2}} = \frac{7}{15} P (X = 1) = P (1 defective bulb) = \frac{^{3} C_{1} \times^{7} C_{1}}{^{10} C_{2}} = \frac{7}{15} P (X = 2) = P (2 defective bulb s) = \frac{^{3} C_{2}}{^{10} C_{2}} = \frac{1}{15}$

Thus, the probability distribution of X is given below,

X	P(X)
0	$\frac{7}{15}$
1	$\frac{7}{15}$
2	$\frac{1}{15}$

Page No 31.15:

Question 28:

Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, then find the probability distribution of X. [NCERT EXEMPLAR]

Answer:

As, four balls are to be drawn without replacement and X denotes the number of red balls drawn

So, X is a random variable that can take values 0, 1, 2, 3 or 4

Now,

$P (X = 0) = P (All white balls) = P (W W W W) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} \times \frac{1}{9} = \frac{1}{495}, P (X = 1) = P (One red balls and three white balls) = P (R W W W) + P (W R W W) + P (W W R W) + P (W W W R) = \frac{8}{12} \times \frac{4}{11} \times \frac{3}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{3}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} \times \frac{8}{9} = 4 \times \frac{8}{495} = \frac{32}{495}, P (X = 2) = P (Two red balls and two white balls) = P (R R W W) + P (R W R W) + P (R W W R) + P (W W R R) + P (W R W R) + P (W R R W) = \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} \times \frac{3}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{7}{10} \times \frac{3}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{7}{10} \times \frac{3}{9} = 6 \times \frac{28}{495} = \frac{168}{495}, P (X = 3) = P (Three red balls and one white ball) = P (R R R W) + P (R R W R) + P (R W R R) + P (W R R R) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{4}{9} + \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} \times \frac{6}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{7}{10} \times \frac{6}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{7}{10} \times \frac{6}{9} = 4 \times \frac{56}{495} = \frac{224}{495}, P (X = 4) = P (All red balls) = P (R R R R) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9} = \frac{70}{495}$

So, the probability distribution of X is as follows:

X	0	1	2	3	4
P(X)	$\frac{1}{495}$	$\frac{32}{495}$	$\frac{168}{495}$	$\frac{224}{495}$	$\frac{70}{495}$

Page No 31.15:

Question 29:

The probability distribution of a random variable X is given below:

X	0	1	2	3
P(X)	k	$\frac{k}{2}$	$\frac{k}{4}$	$\frac{k}{8}$

(i) Determine the value of k

(ii) Determine P(X ≤ 2) and P(X > 2)

(iii) Find P(X ≤ 2) + P(X > 2)

Answer:

We have,

The probability distribution of a random variable X is given below:

X	0	1	2	3
P(X)	k	$\frac{k}{2}$	$\frac{k}{4}$	$\frac{k}{8}$

(i) As, \sum p_{i} = 1 \Rightarrow k + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} = 1 \Rightarrow \frac{8 k + 4 k + 2 k + k}{8} = 1 \Rightarrow \frac{15 k}{8} = 1 ∴ k = \frac{8}{15} (ii) As, P (X \leq 2) = 1 - P (X = 3) = 1 - \frac{k}{8} = 1 - \frac{8}{15 \times 8} = 1 - \frac{1}{15} = \frac{14}{15} Also, P (X > 2) = P (X = 3) = \frac{8}{15 \times 8} = \frac{1}{15} (iii) P (X \leq 2) + P (X > 2) = \frac{14}{15} + \frac{1}{15} [Using (ii)] = \frac{15}{15} = 1

Page No 31.16:

Question 30:

Let, X denote the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that
$P (X = x) = \{\begin{cases} k x & , & if x = 0 or 1 \\ 2 k x & , & if x = 2 \\ k (5 - x) & , & if x = 3 or 4 \\ 0 & , & if x > 4 \end{cases}$
where k is a positive constant. Find the value of k. Also find the probability that you will get admission in (i) exactly one college (ii) at most 2 colleges (iii) at least 2 colleges.

Answer:

The probability distribution of X is

X	0	1	2	3	4
P(X)	0	k	4k	2k	k

The given distribution is a probability distribution.

∴ \sum_{} p_{i} = 1

⇒ 0 + k + 4k + 2k + k = 1
⇒8k = 1
⇒ k = 0.125

(i) P(getting admission in exactly one college) = P(X = 1) = k = 0.125

(ii) P(getting admission in at most 2 colleges) = P( X ≤ 2) = 0 + k + 4k = 5k = 0.625

(iii) P(getting admission in atleast 2 colleges) = P( X ≥ 2) = 4k + 2k + k = 7k = 0.875

Page No 31.42:

Question 1:

Find the mean and standard deviation of each of the following probability distributions:
(i)

x_i :	2	3	4
p_i :	0.2	0.5	0.3

[NCERT EXEMPLAR]
(ii)

x_i :	1	3	4	5
p_i :	0.4	0.1	0.2	0.3

(iii)

x_i :	−5	−4	1	2
p_i :	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{2}$	$\frac{1}{8}$

(iv)

x_i :	−1	0	1	2	3
p_i :	0.3	0.1	0.1	0.3	0.2

(v)

x_i :	1	2	3	4
p_i:	0.4	0.3	0.2	0.1

(vi)

x_i :	0	1	3	5
p_i :	0.2	0.5	0.2	0.1

(vii)

x_i :	−2	−1	0	1	2
p_i :	0.1	0.2	0.4	0.2	0.1

(viii)

x_i :	−3	−1	0	1	3
p_i :	0.05	0.45	0.20	0.25	0.05

(ix)

x_i :	0	1	2	3	4	5
p_i :	$\frac{1}{6}$	$\frac{5}{18}$	$\frac{2}{9}$	$\frac{1}{6}$	$\frac{1}{9}$	$\frac{1}{18}$

[NCERT EXEMPLAR]

Answer:

(i)

x_i	p_i	p_ix_i	p_ix_i²
2	0.2	0.4	0.8
3	0.5	1.5	4.5
4	0.3	1.2	4.8
		$\sum_{}^{}$ p_ix_i = 3.1	$\sum_{}^{}$ p_ix_i² = 10.1

Mean = \sum p_{i} x_{i} = 3.1 Variance = \sum p_{i} {x_{i}}^{2} - {(Mean)}^{2} = 10.1 - {(3.1)}^{2} = 10.1 - 9.61 = 0.49 Step Deviation = \sqrt{Variance} = \sqrt{0.49} = 0.7

(ii)

x_i	p_i	p_ix_i	p_ix_i²
1	0.4	0.4	0.4
3	0.1	0.3	0.9
4	0.2	0.8	3.2
5	0.3	1.5	7.5
		$\sum_{}^{}$ p_ix_i = 3	$\sum_{}^{}$ p_ix_i² = 12

Mean = \sum p_{i} x_{i} = 3 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 12 - 3^{2} = 3 Step Deviation = \sqrt{Variance} = \sqrt{3} = 1.732

(iii)

x_i	p_i	p_ix_i	p_ix_i²
$-$ 5	$\frac{1}{4}$	$- \frac{5}{4}$	$\frac{25}{4}$
$-$ 4	$\frac{1}{8}$	$- \frac{4}{8}$	$\frac{16}{8}$
1	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$\frac{1}{8}$	$\frac{2}{8}$	$\frac{4}{8}$
		$\sum_{}^{}$ p_ix_i = $- 1$	$\sum_{}^{}$ p_ix_i² = $\frac{74}{8}$

Mean = \sum p_{i} x_{i} = - 1 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = \frac{74}{8} - {(- 1)}^{2} = 9.25 - 1 = 8.25 Step Deviation = \sqrt{Variance} = \sqrt{8.25} = 2.872

(iv)

x_i	p_i	p_ix_i	p_ix_i²
$-$ 1	0.3	$-$ 0.3	0.3
0	0.1	0	0
1	0.1	0.1	0.1
2	0.3	0.6	1.2
3	0.2	0.6	1.8
		$\sum_{}^{}$ p_ix_i = 1	$\sum_{}^{}$ p_ix_i² = 3.4

Mean = \sum p_{i} x_{i} = 1 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 3.4 - 1 = 2.4 Step Deviation = \sqrt{Variance} = \sqrt{2.4} = 1.549

(v)

x_i	p_i	p_ix_i	p_ix_i²
1	0.4	0.4	0.4
2	0.3	0.6	1.2
3	0.2	0.6	1.8
4	0.1	0.4	1.6
		$\sum_{}^{}$ p_ix_i = 2	$\sum_{}^{}$ p_ix_i²= 5

Mean = \sum p_{i} x_{i} = 2 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 5 - 2^{2} = 1 Step Deviation = \sqrt{Variance} = \sqrt{1} = 1

(vi)

x_i	p_i	p_ix_i	p_ix_i²
0	0.2	0	0
1	0.5	0.5	0.5
3	0.2	0.6	1.8
5	0.1	0.5	2.5
		$\sum_{}^{}$ p_ix_i = 1.6	$\sum_{}^{}$ p_ix_i²= 4.8

Mean = \sum p_{i} x_{i} = 1.6 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 4.8 - 1 . 6^{2} = 2.24 Step Deviation = \sqrt{Variance} = \sqrt{2.24} = 1.497

(vii)

x_i	p_i	p_ix_i	$\sum_{}^{}$ p_ix_i²
$-$ 2	0.1	$-$ 0.2	0.4
$-$ 1	0.2	$-$ 0.2	0.2
0	0.4	0	0
1	0.2	0.2	0.2
2	0.1	0.2	0.4
		$\sum_{}^{}$ p_ix_i = 0	$\sum_{}^{}$ p_ix_i²=1.2

Mean = \sum p_{i} x_{i} = 0 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 1.2 - 0^{2} = 1.2 Step Deviation = \sqrt{Variance} = \sqrt{1.2} = 1.095

(viii)

x_i	p_i	p_ix_i	p_ix_i_²
$-$ 3	0.05	$-$ 0.15	0.45
$-$ 1	0.45	$-$ 0.45	0.45
0	0.20	0	0
1	0.25	0.25	0.25
3	0.05	0.15	0.45
		$\sum_{}^{}$ p_ix_i = $-$ 0.2	$\sum_{}^{}$ p_ix_i_² =1.6

Mean = \sum p_{i} x_{i} = - 0.2 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 1.6 - {(- 0.2)}^{2} = 1.56 Step Deviation = \sqrt{Variance} = \sqrt{1.56} = 1.249

(ix)

x_i	p_i	p_ix_i	p_ix_i_²
0	$\frac{1}{6}$	0	0
1	$\frac{5}{18}$	$\frac{5}{18}$	$\frac{5}{18}$
2	$\frac{2}{9}$	$\frac{4}{9}$	$\frac{8}{9}$
3	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{2}$
4	$\frac{1}{9}$	$\frac{4}{9}$	$\frac{16}{9}$
5	$\frac{1}{18}$	$\frac{5}{18}$	$\frac{25}{18}$
		$\sum_{}^{}$ p_ix_i = $\frac{35}{18}$	$\sum_{}^{}$ p_ix_i_² = $\frac{35}{6}$

Mean = \sum p_{i} x_{i} = \frac{35}{18} Variance = \sum p_{i} {x_{i}}^{2} - {(Mean)}^{2} = \frac{35}{6} - {(\frac{35}{18})}^{2} = \frac{35}{6} - \frac{1225}{324} = \frac{1890 - 1225}{324} = \frac{665}{324} Step Deviation = \sqrt{Variance} = \sqrt{\frac{665}{324}} = \frac{\sqrt{665}}{18}

Page No 31.43:

Question 2:

A discrete random variable X has the probability distribution given below:

X:	0.5	1	1.5	2
P(X):	k	k²	2k²	k

(i) Find the value of k.

(ii) Determine the mean of the distribution. [NCERT EXEMPLAR]

Answer:

The probability distribution of X is given as:

X:	0.5	1	1.5	2
P(X):	k	k²	2k²	k

(i) As, \sum p_{i} = 1 \Rightarrow k + k^{2} + 2 k^{2} + k = 1 \Rightarrow 3 k^{2} + 2 k - 1 = 0 \Rightarrow 3 k^{2} + 3 k - k - 1 = 0 \Rightarrow 3 k (k + 1) - 1 (k + 1) = 0 \Rightarrow (3 k - 1) (k + 1) = 0 \Rightarrow k = \frac{1}{3} or k = - 1 but k cannot be negative ∴ k = \frac{1}{3} (ii) Mean = \sum p_{i} x_{i} = 0.5 \times k + 1 \times k^{2} + 1.5 \times 2 k^{2} + 2 \times k = 0.5 \times \frac{1}{3} + 1 \times {(\frac{1}{3})}^{2} + 1.5 \times 2 {(\frac{1}{3})}^{2} + 2 \times \frac{1}{3} = \frac{0.5}{3} + \frac{1}{9} + \frac{3}{9} + \frac{2}{3} = \frac{1.5 + 1 + 3 + 6}{9} = \frac{11.5}{9} = \frac{115}{90} = \frac{23}{18}

Page No 31.43:

Question 3:

Find the mean variance and standard deviation of the following probability distribution

x_i:	a	b
p_i :	p	q

where p + q = 1

Answer:

x_i	p_i	p_ix_i	p_ix_i²
a	p	ap	a²p
b	q	bq	b²q
		$\sum_{}^{}$ p_ix_i = ap + bq	$\sum_{}^{}$ p_ix_i²= a²p + b²q

Now, Mean = \sum p_{i} x_{i} = a p + b q Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = a^{2} p + b^{2} q - {(a p + b q)}^{2} = a^{2} p + b^{2} q - a^{2} p^{2} - b^{2} q^{2} - 2 a b p q = a^{2} p - a^{2} p^{2} + b^{2} q - b^{2} q^{2} - 2 a b p q = a^{2} p (1 - p) + b^{2} q (1 - q) - 2 a b p q = a^{2} p q + b^{2} q p - 2 a b p q (∵ p + q = 1) = p q (a^{2} + b^{2} - 2 a b) = p q {(a - b)}^{2} Step Deviation = \sqrt{Variance} = \sqrt{p q {(a - b)}^{2}} = |a - b| \sqrt{p q}

Page No 31.43:

Question 4:

Find the mean and variance of the number of tails in three tosses of a coin. [NCERT EXEMPLAR]

Answer:

Let X denote the number of tails in three tosses of a coin.

Then, X can take the values 0, 1, 2 and 3.

Now,

$P (X = 0) = P (H H H) = \frac{1}{8}, P (X = 1) = P (T H H or H H T or H T H) = \frac{3}{8}, P (X = 2) = P (T T H or T H T or H T T) = \frac{3}{8}, P (X = 3) = P (T T T) = \frac{1}{8}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i²
0	$\frac{1}{8}$	0	0
1	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
2	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
3	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
		$\sum_{}^{}$ p_ix_i = $\frac{3}{2}$	$\sum_{}^{}$ p_ix_i² = 3

Mean = \sum p_{i} x_{i} = \frac{3}{2} Variance = \sum p_{i} {x_{i}}^{2} - {(Mean)}^{2} = 3 - {(\frac{3}{2})}^{2} = 3 - \frac{9}{4} = \frac{3}{4}

Page No 31.43:

Question 5:

Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.

Answer:

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
$P (X = 0) = P (no king) = \frac{^{48} C_{2}}{^{52} C_{2}} = \frac{1128}{1326} = \frac{188}{221} P (X = 1) = P (1 king) = \frac{^{4} C_{1} \times^{48} C_{1}}{^{52} C_{2}} = \frac{192}{1326} = \frac{32}{221} P (X = 2) = P (2 kings) = \frac{^{4} C_{2}}{^{52} C_{2}} = \frac{6}{1326} = \frac{1}{221}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{188}{221}$
1	$\frac{32}{221}$
2	$\frac{1}{221}$

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i²
0	$\frac{188}{221}$	0	0
1	$\frac{32}{221}$	$\frac{32}{221}$	$\frac{32}{221}$
2	$\frac{1}{221}$	$\frac{2}{221}$	$\frac{4}{221}$
		$\sum_{}^{}$ p_ix_i = $\frac{34}{221}$	$\sum_{}^{}$ p_ix_i² = $\frac{36}{221}$

Mean = \sum p_{i} x_{i} = \frac{34}{221} Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = \frac{36}{221} - {(\frac{34}{221})}^{2} = \frac{7956 - 1156}{48841} = \frac{6800}{48841} = \frac{400}{2873}

Page No 31.43:

Question 6:

Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.

Answer:

Let X denote the number of tails in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
$P (X = 0) = P (H H H) = \frac{1}{8}, P (X = 1) = P (T H H or H H T or H T H) = \frac{3}{8} P (X = 2) = P (T T H or T H T or H T T) = \frac{3}{8}, P (X = 3) = P (T T T) = \frac{1}{8}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$

Computation of mean and step deviation

x_i	p_i	p_ix_i	p_ix_i²
0	$\frac{1}{8}$	0	0
1	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
2	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
3	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
		$\sum_{}^{}$ p_ix_i = $\frac{3}{2}$	$\sum_{}^{}$ p_ix_i² = 3

Mean = \sum p_{i} x_{i} = \frac{3}{2} Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 3 - {(\frac{3}{2})}^{2} = \frac{3}{4} Step Deviation = \sqrt{Variance} = \sqrt{\frac{3}{4}} = 0.87

Page No 31.43:

Question 7:

Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.

Answer:

Let X denote the number of bad eggs in a sample of 3 eggs drawn from a lot containing 2 bad eggs and 10 good eggs. Then, X can take the values 0, 1 and 2.
Now,
$P (X = 0) = P (no bad egg) = \frac{^{10} C_{3}}{^{12} C_{3}} = \frac{120}{220} = \frac{6}{11} P (X = 1) = P (1 bad egg) = \frac{^{2} C_{1} \times^{10} C_{2}}{^{12} C_{3}} = \frac{90}{220} = \frac{9}{22} P (X = 2) = P (2 bad eggs) = \frac{^{2} C_{2} \times^{10} C_{1}}{^{12} C_{3}} = \frac{10}{220} = \frac{1}{22}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{6}{11}$
1	$\frac{9}{22}$
2	$\frac{1}{22}$

Computation of mean

x_i	p_i	p_ix_i
0	$\frac{6}{11}$	0
1	$\frac{9}{22}$	$\frac{9}{22}$
2	$\frac{1}{22}$	$\frac{1}{11}$
		$\sum_{}^{}$ p_ix_i = $\frac{1}{2}$

Mean = \sum p_{i} x_{i} = \frac{1}{2}

Page No 31.43:

Question 8:

A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.

Answer:

Let X denote the event of getting twice the number. Then, X can take the values 1, 2, 3, 4, 5 and 6.

Thus, the probability distribution of X is given by

X	P(X)
1	$\frac{11}{36}$
2	$\frac{9}{36}$
3	$\frac{7}{36}$
4	$\frac{5}{36}$
5	$\frac{3}{36}$
6	$\frac{1}{36}$

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i_²
1	$\frac{11}{36}$	$\frac{11}{36}$	$\frac{11}{36}$
2	$\frac{9}{36}$	$\frac{18}{36}$	1
3	$\frac{7}{36}$	$\frac{21}{36}$	$\frac{63}{36}$
4	$\frac{5}{36}$	$\frac{20}{36}$	$\frac{80}{36}$
5	$\frac{3}{36}$	$\frac{15}{36}$	$\frac{75}{36}$
6	$\frac{1}{36}$	$\frac{6}{36}$	1
		$\sum_{}^{}$ p_ix_i = $\frac{91}{36} = 2.5$	$\sum_{}^{}$ p_ix_i_² = $\frac{301}{36} = 8.4$

Mean = \sum p_{i} x_{i} = 2.5 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 8.4 - 6.25 = 2.15

Page No 31.43:

Question 9:

A fair coin is tossed four times. Let X denote the number of heads occurring. Find the probability distribution, mean and variance of X.

Answer:

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTT, HTTT, THHH, ...

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when a coin is tossed 4 times, we can get minimum 0 and maximum 4 strings.)

Now,
$P (X = 0) = P (0 head) = \frac{1}{16} P (X = 1) = P (1 head) = \frac{4}{16} P (X = 2) = P (2 heads) = \frac{6}{16} P (X = 3) = P (3 heads) = \frac{4}{16} P (X = 4) = P (4 heads) = \frac{1}{16}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{16}$
1	$\frac{4}{16}$
2	$\frac{6}{16}$
3	$\frac{4}{16}$
4	$\frac{1}{16}$

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i²
0	$\frac{1}{16}$	0	0
1	$\frac{4}{16}$	$\frac{4}{16}$	$\frac{4}{16}$
2	$\frac{6}{16}$	$\frac{12}{16}$	$\frac{24}{16}$
3	$\frac{4}{16}$	$\frac{12}{16}$	$\frac{36}{16}$
4	$\frac{1}{16}$	$\frac{4}{16}$	1
		$\sum_{}^{}$ p_ix_i = 2	$\sum_{}^{}$ p_ix_i² = 5

Mean = \sum p_{i} x_{i} = 2 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 5 - 4 = 1

Page No 31.43:

Question 10:

A fair die is tossed. Let X denote twice the number appearing. Find the probability distribution, mean and variance of X.

Answer:

Let X denote the event of getting twice the number. Then, X can take the values 2, 4, 6, 8, 10 and 12.

Thus, the probability distribution of X is given by

X	P(X)
2	$\frac{1}{6}$
4	$\frac{1}{6}$
6	$\frac{1}{6}$
8	$\frac{1}{6}$
10	$\frac{1}{6}$
12	$\frac{1}{6}$

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i²
2	$\frac{1}{6}$	$\frac{2}{6}$	$\frac{4}{6}$
4	$\frac{1}{6}$	$\frac{4}{6}$	$\frac{16}{6}$
6	$\frac{1}{6}$	$\frac{6}{6}$	$\frac{36}{6}$
8	$\frac{1}{6}$	$\frac{8}{6}$	$\frac{64}{6}$
10	$\frac{1}{6}$	$\frac{10}{6}$	$\frac{100}{6}$
12	$\frac{1}{6}$	$\frac{12}{6}$	$\frac{144}{6}$
		$\sum_{}^{}$ p_ix_i = 7	$\sum_{}^{}$ p_ix_i² = $\frac{364}{6}$

Mean = \sum p_{i} x_{i} = 7 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 60.7 - 49 = 11.7

Page No 31.43:

Question 11:

A fair die is tossed. Let X denote 1 or 3 according as an odd or an even number appears. Find the probability distribution, mean and variance of X.

Answer:

Let X be 1 for the appearance of odd numbers 1, 3 or 5 on the die. Then,
$P (X = 1) = \frac{3}{6} = \frac{1}{2}$

Let X be 3 for the appearance of even numbers 2, 4 or 6 on the die. Then,
$P (X = 3) = \frac{3}{6} = \frac{1}{2}$

Thus, the probability distribution of X is given by

X	P(X)
1	$\frac{1}{2}$
3	$\frac{1}{2}$

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i²
1	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
3	$\frac{1}{2}$	$\frac{3}{2}$	$\frac{9}{2}$
		$\sum_{}^{}$ p_ix_i = 2	$\sum_{}^{}$ p_ix_i²= 5

Mean = \sum p_{i} x_{i} = 2 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 5 - 4 = 1

Page No 31.43:

Question 12:

A fair coin is tossed four times. Let X denote the longest string of heads occurring. Find the probability distribution, mean and variance of X.

Answer:

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH and TTTT .

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when the coin is tossed 4 times, we can get maximum 4 and minimum 0 strings.)

Now,
$P (X = 0) = P (0 head) = \frac{1}{16} P (X = 1) = P (1 head) = \frac{7}{16} P (X = 2) = P (2 heads) = \frac{5}{16} P (X = 3) = P (3 heads) = \frac{2}{16} P (X = 4) = P (4 heads) = \frac{1}{16}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{16}$
1	$\frac{7}{16}$
2	$\frac{5}{16}$
3	$\frac{2}{16}$
4	$\frac{1}{16}$

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i²
0	$\frac{1}{16}$	0	0
1	$\frac{7}{16}$	$\frac{7}{16}$	$\frac{7}{16}$
2	$\frac{5}{16}$	$\frac{10}{16}$	$\frac{20}{16}$
3	$\frac{2}{16}$	$\frac{6}{16}$	$\frac{18}{16}$
4	$\frac{1}{16}$	$\frac{4}{16}$	1
		$\sum_{}^{}$ p_ix_i = $\frac{27}{16}$	$\sum_{}^{}$ p_ix_i² = $\frac{61}{16}$

Mean = \sum p_{i} x_{i} = \frac{27}{16} = 1.7 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = \frac{61}{16} - \frac{729}{256} = \frac{247}{256} = 0.9

Page No 31.43:

Question 13:

Two cards are selected at random from a box which contains five cards numbered 1, 1, 2, 2, and 3. Let X denote the sum and Y the maximum of the two numbers drawn. Find the probability distribution, mean and variance of X and Y.

Answer:

There are 5 cards numbered 1, 1, 2, 2 and 3.

Let:
X = Sum of two numbers on cards = 2, 3, 4, 5

Y = Maximum of two numbers = 1, 2, 3

Thus, the probability distribution of X is given by

X	P(X)
2	$\frac{1}{10}$
3	$\frac{4}{10}$
4	$\frac{3}{10}$
5	$\frac{2}{10}$

Total number of cards = 2 cards with "1" on them + 2 cards with "2" on them + 1 card with "3" on it
= 5

Total number of possible choices (in drawing the two cards)= Number of ways in which the two cards can be drawn from the total 5
                                                                                                 =⁵C₂
                                                                                                 = $\frac{5!}{2! 3!}$
                                                                                                 = 10

Probabilty that the two cards drawn would be having the numbers 1 and 1 on them is P(11).

P(11) = $\frac{{}^{2}C_{2} \times {}^{2}C_{0} \times {}^{1}C_{0}}{{}^{5}C_{2}}$
$= \frac{1 \times 1 \times 1}{10} = \frac{1}{10}$

Probabilty that the two cards drawn would be having the numbers 1 and 2 on them is P(11).

P(11) = $\frac{{}^{2}C_{2} \times {}^{2}C_{0} \times {}^{1}C_{0}}{{}^{5}C_{2}}$
$= \frac{1 \times 1 \times 1}{10} = \frac{1}{10}$

"1" and "2" on them

⇒ P(12)

²C₁ × ²C₁ × ¹C₀

⁵C₂

	"1's"	"2's"	"3's"	Total
Available	2	2	1	5
To Choose	1	1	0	2
Choices	²C₁	²C₁	¹C₀	⁵C₂

× 1

2 × 2 × 1

"1" and "3" on them

⇒ P(13)

²C₁ × ²C₀ × ¹C₁

⁵C₂

	"1's"	"2's"	"3's"	Total
Available	2	2	1	5
To Choose	1	0	1	2
Choices	²C₁	²C₀	¹C₁	⁵C₂

× 1 ×

2 × 1 × 1

"2" and "2" on them

⇒ P(22)

²C₀ × ²C₂ × ¹C₁

⁵C₂

	"1's"	"2's"	"3's"	Total
Available	2	2	1	5
To Choose	0	2	0	2
Choices	²C₀	²C₂	¹C₀	⁵C₂

1 × 1 × 1

"2" and "3" on them

⇒ P(23)

²C₀ × ²C₁ × ¹C₁

⁵C₂

	"1's"	"2's"	"3's"	Total
Available	2	2	1	5
To Choose	0	1	1	2
Choices	²C₀	²C₁	¹C₁	⁵C₂

1 ×

× 1

1 × 2 × 1

Probability for the sum of the numbers on the cards drawn to be

2 ⇒ P(X = 2)

P(11)

3 ⇒ P(X = 3)

P(12)

4 ⇒ P(X = 4)

P(13 or 22) i.e. P(13 ∪ 22)

P(13) + P(22)

5 ⇒ P(X = 2)

P(23)

The probabilty distribution of "x" would be

P(X = x)

Calculations for Mean and Standard Deviations

P (X = x)

px
[x × P (X = x)]

x²

px²
[x² × P (X = x)]

Total

138

= 3.6

= 13.8

The Expected Value of the sum

⇒ Expectation of "x"

⇒ E (x)	=	Σ px
	=	3.6

Variance of the sum of the numbers on the cards

⇒ var (x)	=	E (x²) − (E(x))²
⇒ var (x)	=	Σ px² − (Σ px)²
	=	13.8 − (3.6)²
	=	13.8 − 12.96
	=	0.84

Standard Deviation of the sum of the numbers on the cards

⇒ SD (x)	=	+ √ Var (x)
	=	+ √ 0.84
	=	+ 0.917

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i_²
2	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{4}{10}$
3	$\frac{4}{10}$	$\frac{12}{10}$	$\frac{36}{10}$
4	$\frac{3}{10}$	$\frac{12}{10}$	$\frac{48}{10}$
5	$\frac{2}{10}$	1	$\frac{50}{10}$
		$\sum_{}^{}$ p_ix_i = $\frac{36}{10} = 3.6$	$\sum_{}^{}$ p_ix_i_² = $\frac{138}{10} = 13.8$

Mean = \sum p_{i} x_{i} = 3.6 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 13.8 - 12.96 = 0.84

Thus, the probability distribution of Y is given by

Y	P(Y)
1	0.1
2	0.5
3	0.4

Computation of mean and variance

y_i	p_i	p_iy_i	p_iy_i_²
1	0.1	0.1	0.1
2	0.5	1	2
3	0.4	1.2	3.6
		$\sum_{}^{}$ p_ix_i = 2.3	$\sum_{}^{}$ p_ix_i_² = 5.7

Mean = \sum p_{i} y_{i} = 2.3 Variance = {\sum p_{i} {y_{i}}^{2}}_{}^{} - {(Mean)}^{2} = 5.7 - 5.29 = 0.41

Page No 31.43:

Question 14:

A die is tossed twice. A 'success' is getting an odd number on a toss. Find the variance of the number of successes.

Answer:

It is given that "success" denotes the event of getting the numbers 1, 3 or 5. Then,
$P (success) = \frac{1}{2}$

Also, "failure" denotes the event of getting the numbers 2, 4 or 6. Then,
$P (failure) = \frac{1}{2}$

Let X denote the event of getting success.Then, X can take the values 0, 1 and 2.
Now,
$P (X = 0) = P (no success) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} P (X = 1) = P (1 success) = (\frac{1}{2} \times \frac{1}{2}) + (\frac{1}{2} \times \frac{1}{2}) = \frac{1}{2} P (X = 2) = P (2 success) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{4}$
1	$\frac{1}{2}$
2	$\frac{1}{4}$

Computation of mean and variance

x_i	p_i	p_ix_i	p_ix_i_²
0	$\frac{1}{4}$	0	0
1	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$\frac{1}{4}$	$\frac{1}{2}$	1
		$\sum_{}^{}$ p_ix_i = 1	$\sum_{}^{}$ p_ix_i_² = $\frac{3}{2}$

Mean = \sum p_{i} x_{i} = 1 Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = \frac{3}{2} - 1 = \frac{1}{2}

Page No 31.43:

Question 15:

A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability distribution of the number of defective bulbs.

Answer:

Let X denote the number of defective bulbs in a sample of 3 bulbs drawn from a bag containing 13 bulbs. 5 bulbs in the bag turn out to be defective. Then, X can take the values 0, 1, 2 and 3.

Now,
$P (X = 0) = P (no defective bulb) = \frac{^{8} C_{3}}{^{13} C_{3}} = \frac{56}{286} = \frac{28}{143} P (X = 1) = P (1 defective bulb) = \frac{^{5} C_{1} \times^{8} C_{2}}{^{13} C_{3}} = \frac{140}{286} = \frac{70}{143} P (X = 2) = P (2 defective bulbs) = \frac{^{5} C_{2} \times^{8} C_{1}}{^{13} C_{3}} = \frac{80}{286} = \frac{40}{143} P (X = 3) = P (3 defective bulbs) = \frac{^{5} C_{3}}{^{13} C_{3}} = \frac{10}{286} = \frac{5}{143}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{28}{143}$
1	$\frac{70}{143}$
2	$\frac{40}{143}$
3	$\frac{5}{143}$

Page No 31.43:

Question 16:

In roulette, Figure, the wheel has 13 numbers 0, 1, 2, ...., 12 marked on equally spaced slots. A player sets Rs 10 on a given number. He receives Rs 100 from the organiser of the game if the ball comes to rest in this slot; otherwise he gets nothing. If X denotes the player's net gain/loss, find E (X).
Figure

Answer:

The wheel has 13 numbers, i.e. 0, 1, 2, ... ,12 marked on equally spaced slots.

∴ Probability of ball resting on any particular number = $\frac{1}{13}$

Let the player set Rs 10 on a given number k.

P(player receives Rs 100) = P(ball rests on it) = $\frac{1}{13}$

X denotes the player's net gain or loss. If he gets the required number, then his gain is Rs 90 (100-10).

If the ball does not rest on the number, then it rests on any of the other 12 numbers. In that case, the player's loss is Rs 10.

Thus, the probability distribution of X is given by

Computation of mean

x_i	p_i	p_ix_i
90	$\frac{1}{13}$	$\frac{90}{13}$
$-$ 10	$\frac{12}{13}$	$\frac{- 120}{13}$
		$\sum_{}^{}$ p_ix_i = $\frac{- 30}{13}$

Mean = E (x) = \sum p_{i} x_{i} = - \frac{30}{13}

Page No 31.44:

Question 17:

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence, find the mean of the distribution. [CBSE 2014]

Answer:

Let X denotes the number of red cards drawn.

Then, X can take the values 0, 1, 2 or 3.

Now,

$P (X = 0) = P (B B B) = \frac{26}{52} \times \frac{25}{51} \times \frac{24}{50} = \frac{2}{17}, P (X = 1) = P (R B B or B R B or B B R) = 3 \times \frac{26}{52} \times \frac{26}{51} \times \frac{25}{50} = \frac{13}{34}, P (X = 2) = P (R R B or R B R or B R R) = 3 \times \frac{26}{52} \times \frac{25}{51} \times \frac{26}{50} = \frac{13}{34}, P (X = 3) = P (R R R) = \frac{26}{52} \times \frac{25}{51} \times \frac{24}{50} = \frac{2}{17}$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{2}{17}$
1	$\frac{13}{34}$
2	$\frac{13}{34}$
3	$\frac{2}{17}$

Mean = \sum p_{i} x_{i} = 0 \times \frac{2}{17} + 1 \times \frac{13}{34} + 2 \times \frac{13}{34} + 3 \times \frac{2}{17} = 0 + \frac{13}{34} + \frac{26}{34} + \frac{6}{17} = \frac{51}{34} = \frac{3}{2} = 1.5

Disclaimer: The answer given in the book is incorrect. The same has been corrected here.

Page No 31.44:

Question 18:

An urn contains 5 red and 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, then find the mean and variance of X. [CBSE 2015]

Answer:

As, X represent the number of black balls drawn.

So, it can take values 0, 1 and 2. Yes, X is a random variable.

Now,

$P (X = 0) = P (R R) = \frac{5}{7} \times \frac{4}{6} = \frac{10}{21}, P (X = 1) = P (R B or B R) = 2 \times \frac{5}{7} \times \frac{2}{6} = \frac{10}{21}, P (X = 2) = P (B B) = \frac{2}{7} \times \frac{1}{6} = \frac{1}{21} Mean = \sum p_{i} x_{i} = 0 \times \frac{10}{21} + 1 \times \frac{10}{21} + 2 \times \frac{1}{21} = \frac{10}{21} + \frac{2}{21} = \frac{12}{21} = \frac{4}{7} Also, \sum p_{i} {x_{i}}^{2} = 0^{2} \times \frac{10}{21} + 1^{2} \times \frac{10}{21} + 2^{2} \times \frac{1}{21} = \frac{10}{21} + \frac{4}{21} = \frac{14}{21} = \frac{2}{3} So, variance = \sum p_{i} {x_{i}}^{2} - {(Mean)}^{2} = \frac{2}{3} - {(\frac{4}{7})}^{2} = \frac{2}{3} - \frac{16}{49} = \frac{98 - 48}{147} = \frac{50}{147}$

Page No 31.44:

Question 19:

Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X. [CBSE 2015]

Answer:

As, X denote the larger of the two numbers obtained.

So, X can take the values 3, 4, 5, 6 and 7.

Now,

$P (X = 3) = P \{(2, 3) or (3, 2)\} = 2 \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{15}, P (X = 4) = P \{(2, 4) or (4, 2) or (3, 4) or (4, 3)\} = 4 \times \frac{1}{6} \times \frac{1}{5} = \frac{2}{15}, P (X = 5) = P \{(2, 5) or (5, 2) or (3, 5) or (5, 3) or (4, 5) or (5, 4)\} = 6 \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{5}, P (X = 6) = P \{(2, 6) or (6, 2) or (3, 6) or (6, 3) or (4, 6) or (6, 4) or (5, 6) or (6, 5)\} = 8 \times \frac{1}{6} \times \frac{1}{5} = \frac{4}{15}, P (X = 7) = P \{(2, 7) or (7, 2) or (3, 7) or (7, 3) or (4, 7) or (7, 4) or (5, 7) or (7, 5) or (6, 7) or (7, 6)\} = 10 \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{3}$

The probability distribution of X is as follows:

X:	3	4	5	6	7
P(X):	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{4}{15}$	$\frac{1}{3}$

So, Mean, E (X) = 3 \times \frac{1}{15} + 4 \times \frac{2}{15} + 5 \times \frac{1}{5} + 6 \times \frac{4}{15} + 7 \times \frac{1}{3} = \frac{3}{15} + \frac{8}{15} + \frac{15}{15} + \frac{24}{15} + \frac{35}{15} = \frac{85}{15} = \frac{17}{3} Also, E (X^{2}) = 3^{2} \times \frac{1}{15} + 4^{2} \times \frac{2}{15} + 5^{2} \times \frac{1}{5} + 6^{2} \times \frac{4}{15} + 7^{2} \times \frac{1}{3} = \frac{9}{15} + \frac{32}{15} + \frac{75}{15} + \frac{144}{15} + \frac{245}{15} = \frac{505}{15} = \frac{101}{3} Var (X) = E (X^{2}) - {[E (X)]}^{2} = \frac{101}{3} - {(\frac{17}{3})}^{2} = \frac{101}{3} - \frac{289}{9} = \frac{303 - 289}{9} = \frac{14}{9}

Page No 31.44:

Question 20:

In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown. The man decided to thrown a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses.

Answer:

The man may get number greater than 4 in the first throw and then he quits the game. He may get a number less than equatl to 4 in the first throw and in the second throw he may get the number greater than 4 and quits the game.
In the first two throws he gets a number less than equal to 4 and in the third throw he may get a number greater than 6. He may not get number greater than 4 in any one of three throws.
Let X be the amount he wins/looses.
Then, X can take values -3, 3, 4, 5 such that
P (X = 5) = P(Getting number greater than 4 in first throw) = $\frac{1}{3}$
P (X = 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) = $\frac{4}{6} \times \frac{2}{6} = \frac{2}{9}$
P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) = $\frac{4}{6} \times \frac{4}{6} \times \frac{2}{6} = \frac{4}{27}$
P (X = -3) = P(Getting number less than equal to 4 in all three throws) = $\frac{4}{6} \times \frac{4}{6} \times \frac{4}{6} = \frac{8}{27}$

X	5	4	3	-3
P(X)	$\frac{1}{3}$	$\frac{2}{9}$	$\frac{4}{27}$	$\frac{8}{27}$

E (X) = (5 \times \frac{1}{3}) + 4 (\frac{2}{9}) + 3 (\frac{4}{27}) - 3 (\frac{8}{27}) = \frac{1}{27} (45 + 24 + 12 - 24) = \frac{57}{27}

Expected value of the amount he wins/loses is

\frac{57}{27}

Page No 31.45:

Question 1:

If a random variable X has the following probability distribution:

X :	0	1	2	3	4	5	6	7	8
P (X) :	a	3a	5a	7a	9a	11a	13a	15a	17a

then the value of a is

(a)

\frac{7}{81}

(b)

\frac{5}{81}

(c)

\frac{2}{81}

(d)

\frac{1}{81}

Answer:

(d) $\frac{1}{81}$

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

$\Rightarrow a + 3 a + 5 a + 7 a + 9 a + 11 a + 13 a + 15 a + 17 a = 1 \Rightarrow 81 a = 1 \Rightarrow a = \frac{1}{81}$

Page No 31.45:

Question 2:

A random variable X has the following probability distribution:

X :	1	2	3	4	5	6	7	8
P (X) :	0.15	0.23	0.12	0.10	0.20	0.08	0.07	0.05

For the events E = {X : X is a prime number}, F = {X : X < 4}, the probability P (E ∪ F) is

(a) 0.50
(b) 0.77
(c) 0.35
(d) 0.87

Answer:

(b) 0.77

E = {X is a prime number} = {2, 3, 5, 7}

$P (E) = P (2) + P (3) + P (5) + P (7) = 0.62 F = {X < 4} = {X = 1, 2, 3} P (F) = P (1) + P (2) + P (3) = 0.5 Now, E \cap F = {2, 3} P (E \cap F) = P (2) + P (3) = 0.35 P (E \cup F) = P (E) + P (F) - P (E \cap F) = 0.62 + 0.50 - 0.35 = 0.77$

Page No 31.46:

Question 3:

A random variable X takes the values 0, 1, 2, 3 and its mean is 1.3. If P (X = 3) = 2 P (X = 1) and P (X = 2) = 0.3, then P (X = 0) is
(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.4

Answer:

(d) 0.4

Let:
P(X = 0) = m
P(X = 1) = k.

Now,
P(X = 3) = 2k

x_i	p_i	p_ix_i
0	m	0
1	k	k
2	0.3	0.6
3	2k	6k

Mean =

\sum_{}^{}

p_ix_i

0 + k + 0.6 + 6 k = 1.3 \Rightarrow 7 k = 1.3 - 0.6 \Rightarrow k = \frac{0.7}{7} = 0.1

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

\Rightarrow m + 0.1 + 0.3 + 0.2 = 1 \Rightarrow m + 0.6 = 1 \Rightarrow m = 0.4

Page No 31.46:

Question 4:

A random variable has the following probability distribution:

X = x_i :	0	1	2	3	4	5	6	7
P (X = x_i) :	0	2 p	2 p	3 p	p²	2 p²	7 p²	2 p

The value of p is

(a) 1/10
(b) −1
(c) −1/10
(d) 1/5

Answer:

(a) 1/10

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

$\Rightarrow 0 + 2 p + 2 p + 3 p + p^{2} + 2 p^{2} + 7 p^{2} + 2 p = 1 \Rightarrow 10 p^{2} + 9 p - 1 = 0 \Rightarrow (10 p - 1) (p + 1) = 0 \Rightarrow p = \frac{1}{10} or - 1 (Neglecting - 1 as the value of the probabilitiy cannot be negative)$

Page No 31.46:

Question 5:

If X is a random-variable with probability distribution as given below:

X = x_i :	0	1	2	3
P (X = x_i) :	k	3 k	3 k	k

The value of k and its variance are

(a) 1/8, 22/27
(b) 1/8, 23/27
(c) 1/8, 24/27
(d) 1/8, 3/4

Answer:

(d) 1/8, 3/4

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

$\Rightarrow k + 3 k + 3 k + k = 1 \Rightarrow 8 k = 1 \Rightarrow k = \frac{1}{8}$

Now,

x_i	p_i	p_ix_i	p_ix_i_²
0	k = $\frac{1}{8}$	0	0
1	3k = $\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
2	3k = $\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
3	k = $\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
		$\sum_{}^{}$ p_ix_i = $\frac{12}{8} = \frac{3}{2}$	$\sum_{}^{}$ p_ix_i_² = $\frac{24}{8}$

Mean = \sum p_{i} x_{i} = \frac{3}{2} Variance = {\sum p_{i} {x_{i}}^{2}}_{}^{} - {(Mean)}^{2} = \frac{24}{8} - {(\frac{3}{2})}^{2} = \frac{24}{8} - \frac{9}{4} = \frac{24 - 18}{8} = \frac{6}{8} = \frac{3}{4}

Page No 31.46:

Question 6:

Mark the correct alternative in the following question:

The probability distribution of a discrete random variable X is given below:

X:	2	3	4	5
P(X):	$\frac{5}{k}$	$\frac{7}{k}$	$\frac{9}{k}$	$\frac{11}{k}$

The value of k is

(a) 8 (b) 16 (c) 32 (d) 48

Answer:

The probability distribution of a discrete random variable X is given below:

X:	2	3	4	5
P(X):	$\frac{5}{k}$	$\frac{7}{k}$	$\frac{9}{k}$	$\frac{11}{k}$

As, \sum p_{i} = 1 \Rightarrow \frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1 \Rightarrow \frac{32}{k} = 1 ∴ k = 32

Disclaimer: The question is incorrect, it should be "k" instead "E(X)". The same has been corrected here.

Page No 31.46:

Question 7:

Mark the correct alternative in the following question:

For the following probability distribution:

X:	−4	−3	−2	−1	0
P(X):	0.1	0.2	0.3	0.2	0.2

The value of E(X) is

(a) 0 (b) −1 (c) −2 (d) −1.8

Answer:

The probability distribution of X is given below:

X:	−4	−3	−2	−1	0
P(X):	0.1	0.2	0.3	0.2	0.2

E (X) = (- 4) \times 0.1 + (- 3) \times 0.2 + (- 2) \times 0.3 + (- 1) \times 0.2 + 0 \times 0.2 = - 0.4 - 0.6 - 0.6 - 0.2 = - 1.8

Hence, the correct alternative is option (d).

Page No 31.46:

Question 8:

Mark the correct alternative in the following question:

For the following probability distribution:

X:	1	2	3	4
P(X):	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

The value of E(X²) is

(a) 3 (b) 5 (c) 7 (d) 10

Answer:

The probability distribution of X is given below:

X:	1	2	3	4
P(X):	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

E (X^{2}) = 1^{2} \times \frac{1}{10} + 2^{2} \times \frac{1}{5} + 3^{2} \times \frac{3}{10} + 4^{2} \times \frac{2}{5} = \frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10} = \frac{100}{10} = 10

Hence, the correct alternative is option (d).

Page No 31.46:

Question 9:

Mark the correct alternative in the following question:

Let X be a discrete random variable. Then the variance of X is

(a) E(X²) (b) E(X²) + (E(X))² (c) E(X²) $-$ (E(X))² (d) $\sqrt{E (X^{2}) - {(E (X))}^{2}}$

Answer:

Since, the variance of a discrete random variable X is given by:

Var(X) = E(X²) $-$ (E(X))²

Hence, the correct alternative is option (c).

Page No 31.46:

Question 10:

Let X be a discrete random variable. The probability distribution of X is given below:

X:	30	10	–10
P(X):	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{1}{2}$

Then E(X) is equal to
(a) 6
(b) 4
(c) 3
(d) –5

Answer:

For a discrete random variable X,

X	P(X)
30	$\frac{1}{5}$
10	$\frac{3}{10}$
–10	$\frac{1}{2}$

Since E (X) = \sum_{i = 1}^{n} x_{i} P (x_{i}) = 30 \times \frac{1}{5} + 10 \times \frac{3}{10} + (- 10) \times \frac{1}{2} = 6 + 3 - 5 = 9 - 5 i . e . E (X) = 4

Hence, the correct answer is option B.

Page No 31.47:

Question 1:

If X is a random variable with the following probability distribubtion:
X: x₁ x₂ ________ x_n
P(X): p₁ p₂ ________ p_n
Then, Mean (X) = _____________.

Answer:

For a given random variable X,

X	P(X)
x₁	p₁
x₂	p₂
-	-
-	-
x_n	p_n

then Mean (X) = x₁p₁ + x₂p₂ + ......... + x_np_n

\begin{array}{rcl} = & \sum_{i = 1}^{n} x_{i} p_{i} \\ Mean (X) & = & \sum_{i = 1}^{n} p_{i} x_{i} \end{array}

Page No 31.47:

Question 2:

In Q. No. 1, Var (X) = ____________.

Answer:

Var (X) = E (X²) − (E(X))²
Here,

X	P(X)	X²
x₁	P₁	$x_{1}^{2}$
x₂	P₂	$x_{2}^{2}$
-	-	-
-	-	-
x_n	P_n	$x_{n}^{2}$

\begin{array}{rcl} Since E (x) & = & \sum_{i = 1}^{n} p_{i} x_{i} \\ E (x^{2}) & = & \sum_{i = 1}^{n} p_{i} {x_{i}}^{2} \end{array}

∴ var X = E(X²) − (E(X))²
i.e. var X =

\sum_{i = 1}^{n} p_{i} x i^{2} - {(\sum_{i = 1}^{n} p_{i} x_{i})}^{2}

Page No 31.47:

Question 3:

A discrete random variable X has the following probability distribution:
X: 1 2 3 4 5 6 7
P(X): c 2c 2c 3c c² 2c² 7c²+c
Then, P(X ≤ 2) = ____________.

Answer:

For a discrete random variable X

X	P(X)
1	c
2	2c
3	2c
4	3c
5	c²
6	2c²
7	7c²+ c

$Since \sum_{i = 1}^{7} p (x_{i}) = 1 i . e . c + 2 c + 2 c + 3 c + c^{2} + 2 c^{2} + 7 c^{2} + c = 1 i . e . 8 c + c + 10 c^{2} = 1 i . e . 10 c^{2} + 9 c - 1 = 0 i . e . 10 c^{2} + 10 c - c - 1 = 0 i . e . 10 c (c + 1) - 1 (c + 1) = 0 i . e . c = \frac{1}{10}$
$∴ P (X \leq 2) = c + 2 c = 2 c + c = 3 c = 3 \times \frac{1}{10} i . e P (X \leq 2) = \frac{3}{10}$

Page No 31.47:

Question 4:

If a discrete random variable X has the following probability distribution:
X: $\frac{1}{2} 1 \frac{3}{2} 2$
P(X): c c² 2c² c
Then, c = ____________.

Answer:

For a discrete random variable,
Since sum of probability is 1
∴ c + c² + 2c² + c = 1
i.e. 3c² + 2c = 1
i.e. 3c² + 2c − 1 = 0
i.e. 3c² + 3c − c − 1 = 0
i.e. 3c (c + 1) − 1 (c + 1) = 0
i.e. c = $\frac{1}{3}$ (∵ c ≠ −1)

Page No 31.47:

Question 5:

If a random variable X has the following probability distribution:
X: 0 1 2 3
P(X): $\frac{1}{5} \frac{3}{10} \frac{2}{5} \frac{1}{10}$
Then, E(X²) =______________.

Answer:

For random variable X

X	P(X)	X²
0	$\frac{1}{5}$	0
1	$\frac{3}{10}$	1
2	$\frac{2}{5}$	4
3	$\frac{1}{10}$	9

$\begin{array}{rcl} E (X^{2}) & = & \sum_{i = 1}^{4} x^{2} p (x) \\ = & 0 \times \frac{1}{5} + 1 \times \frac{3}{10} + 4 \times \frac{2}{5} + 9 \times \frac{1}{10} \\ = & \frac{3}{10} + \frac{8}{5} + \frac{9}{10} \\ = & \frac{3 + 16 + 9}{10} \\ = & \frac{28}{10} \end{array}$

i . e E (X^{2}) = \frac{14}{5}

Page No 31.47:

Question 1:

Write the values of 'a' for which the following distribution of probabilities becomes a probability distribution:

X = x_i :	−2	−1	0	1
P (X = x_i) :	$\frac{1 - a}{4}$	$\frac{1 + 2 a}{4}$	$\frac{1 - 2 a}{4}$	$\frac{1 + a}{4}$

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = $-$ 2) + P (X = $-$ 1) + P (X = 0) + P (X = 1) = 1

$\Rightarrow \frac{1 - a}{4} + \frac{1 + 2 a}{4} + \frac{1 - 2 a}{4} + \frac{1 + a}{4} = 1 \Rightarrow \frac{4}{4} = 1 \Rightarrow 1 = 1 Now, 0 \leq \frac{1 - a}{4} \leq 1 \Rightarrow 0 \leq 1 - a \leq 4 \Rightarrow - 1 \leq - a \leq 3 \Rightarrow - 3 \leq a \leq 1 . . . (1) 0 \leq \frac{1 + a}{4} \leq 1 \Rightarrow 0 \leq 1 + a \leq 4 \Rightarrow - 1 \leq a \leq 3 . . . (2) 0 \leq \frac{1 - 2 a}{4} \leq 1 \Rightarrow 0 \leq 1 - 2 a \leq 4 \Rightarrow - 1 \leq - 2 a \leq 3 \Rightarrow - \frac{3}{2} \leq a \leq \frac{1}{2} . . . (3) 0 \leq \frac{1 + 2 a}{4} \leq 1 \Rightarrow 0 \leq 1 + 2 a \leq 4 \Rightarrow - 1 \leq 2 a \leq 3 \Rightarrow - \frac{1}{2} \leq a \leq \frac{3}{2} . . . (4) From (1), (2), (3) and (4), we get - \frac{1}{2} \leq a \leq \frac{1}{2}$

Page No 31.47:

Question 2:

For what value of k the following distribution is a probability distribution?

X = x_i :	0	1	2	3
P (X = x_i) :	2k⁴	3k² − 5k³	2k − 3k²	3k − 1

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

$\Rightarrow 2 k^{4} + 3 k^{2} - 5 k^{3} + 2 k - 3 k^{2} + 3 k - 1 = 1 \Rightarrow 2 k^{4} - 5 k^{3} + 5 k = 2 \Rightarrow 2 k^{4} - 5 k^{3} + 5 k - 2 = 0 \Rightarrow (k - 1) (k - 2) (2 k^{2} + k - 1) = 0 \Rightarrow (k - 1) (k - 2) (2 k - 1) (k + 1) = 0 \Rightarrow k = - 1, \frac{1}{2}, 1, 2 (Neglecting - 1, 1 and 2 as they give the value of probability negative or greater than 1)$

∴ k = $\frac{1}{2}$

Page No 31.48:

Question 3:

If X denotes the number on the upper face of a cubical die when it is thrown, find the mean of X.

Answer:

A cubical die can show 1, 2, 3, 4, 5 or 6 on its face.

x_i	p_i	p_ix_i
1	$\frac{1}{6}$	$\frac{1}{6}$
2	$\frac{1}{6}$	$\frac{2}{6}$
3	$\frac{1}{6}$	$\frac{3}{6}$
4	$\frac{1}{6}$	$\frac{4}{6}$
5	$\frac{1}{6}$	$\frac{5}{6}$
6	$\frac{1}{6}$	$\frac{6}{6}$

Mean =

\sum_{}^{}

p_ix_i =

\frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} = \frac{21}{6} = 3.5

Page No 31.48:

Question 4:

If the probability distribution of a random variable X is given by

X = x_i :	1	2	3	4
P (X = x_i) :	2k	4k	3k	k

Write the value of k.

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

$\Rightarrow 2 k + 4 k + 3 k + k = 1 \Rightarrow 10 k = 1 \Rightarrow k = \frac{1}{10} = 0.1$

Page No 31.48:

Question 5:

Find the mean of the following probability distribution:

X = x_i :	1	2	3
P (X = x_i) :	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{5}{8}$

Answer:

x_i	p_i	p_ix_i
1	$\frac{1}{4}$	$\frac{1}{4}$
2	$\frac{1}{8}$	$\frac{2}{8}$
3	$\frac{5}{8}$	$\frac{15}{8}$

Mean =

\sum_{}^{}

p_ix_i =

\frac{1}{4} + \frac{2}{8} + \frac{15}{8} = \frac{2 + 2 + 15}{8} = \frac{19}{8}

Page No 31.48:

Question 6:

If the probability distribution of a random variable X is as given below:

X = x_i :	1	2	3	4
P (X = x_i) :	c	2c	4c	4c

Write the value of P (X ≤ 2).

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

$\Rightarrow c + 2 c + 4 c + 4 c = 1 \Rightarrow 11 c = 1 \Rightarrow c = \frac{1}{11} Now, P (X \leq 2) = P (X = 1) + P (X = 2) = \frac{1}{10} + \frac{2}{10} = \frac{3}{11}$

Page No 31.48:

Question 7:

A random variable has the following probability distribution:

X = x_i :	1	2	3	4
P (X = x_i) :	k	2k	3k	4k

Write the value of P (X ≥ 3).

Answer:

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

$\Rightarrow k + 2 k + 3 k + 4 k = 1 \Rightarrow 10 k = 1 \Rightarrow k = \frac{1}{10} Now, P (X \geq 3) = P (X = 3) + P (X = 4) = \frac{3}{10} + \frac{4}{10} = \frac{7}{10}$

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