Question 1:

Which of the following cannot be valid assignment of probability for elementary events or outcomes of sample space S = {w₁, w₂, w₃, w₄, w₅, w₆, w₇}:

Elementary events:	w₁	w₂	w₃	w₄	w₅	w₆	w₇
(i)	0.1	0.01	0.05	0.03	0.01	0.2	0.6
(ii)	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$
(iii)	0.7	0.6	0.5	0.4	0.3	0.2	0.1
(iv)	$\frac{1}{14}$	$\frac{2}{14}$	$\frac{3}{14}$	$\frac{4}{14}$	$\frac{5}{14}$	$\frac{6}{14}$	$\frac{15}{14}$

Answer:

(i)

Here, each of the numbers p(ω_i) is positive and less than 1.
∴ Sum of probabilities = $p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})$
                                    = 0. 1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Thus, the assignment is valid.

(ii)

     Here, each of the numbers p(ω_i) is positive and less than 1.
    ∴ Sum of probabilities = $p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})$
                                       = $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = 7 \times \frac{1}{7} = 1$
     Thus, the assignment is valid.

(iii)

        Here, each of the numbers p(ω_i) is positive and less than 1.
   ∴ Sum of probabilities = $p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})$
                                         = 0.7 + 0.6 + 0.5 + 0.4 + 0.3 + 0.2 + 0.1
                                         = 2.8 ≠ 1
       Thus, the assignment is not valid.

(iv)

      Given:
       $p (ω_{7}) = \frac{15}{14} > 1$
       Thus, the assignment is not valid.

Page No 33.65:

Question 1:

(a) If A and B be mutually exclusive events associated with a random experiment such that P(A) = 0.4 and P(B) = 0.5, then find:
(i) P (A ∪ B)
(ii) $P (A \cap B)$
(iii) P ( $A$ ∩ B)
(iv) P (A ∩ $B$ ).

(b) A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35.
Find (i) P (A ∪ B)
(ii) $P (A \cap B)$
(iii) P (A ∩ $B$ )
(iv) P (B ∩ $A$ )

(c) Fill in the blanks in the following table:

	P (A)	P (B)	P (A ∩ B)	P (A ∪ B)
(i)	$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	......
(ii)	0.35	....	0.25	0.6
(iii)	0.5	0.35	.....	0.7

Answer:

$P (\bar{A} \cap B) = 1 - P (A \cup B)$
$\Rightarrow P (A \cup B) = 1 - P (\bar{A} \cap B) = 1 - \frac{1}{3} = \frac{2}{3}$
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
∴ P(A) + P (B) = P (A ∪ B) + P (A ∩ B)
                         = $\frac{2}{3} + \frac{1}{6} = \frac{4 + 1}{6} = \frac{5}{6}$
Thus, P(A) + P (B) = $\frac{5}{6}$           ...(i)
Again, P (A ∩ B) = P(A) × P(A) = $\frac{1}{6}$
By formula, we have:
{P(A) − P (B)}² = {P(A) + P (B)}² − 4 × P(A) × P(B)
                          = ${(\frac{5}{6})}^{2} - \frac{4}{6} = \frac{25}{36} - \frac{4}{6} = \frac{25 - 24}{36} = \frac{1}{36}$
∴ P(A) − P(B) = $\frac{1}{6}$           ...(ii)
From (i) and (ii), we get:
2P(A) = 1
Hence, P(A) = $\frac{1}{2}$ and P(B) = $\frac{1}{3}$

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