Rd Sharma 2020 2021 for Class 7 Maths Chapter 25 - Data Handling Iv Probability

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Page No 25.6:

Question 1:

A coin is tossed 1000 times with the following frequencies:

Head: 445,

Tail: 555

When a coin is tossed at random, what is the probability of getting (i) a head? (ii) a tail?

Answer:

Total number of times a coin is tossed = 1000
Number of times a head comes up = 445
Number of times a tail comes up = 555

(i) Probability of getting a head = $\frac{Number of heads}{Total no . of trails} = \frac{445}{1000} = 0.445$

(ii) Probability of getting a tail = $\frac{Number of tails}{Total no . of trails} = \frac{555}{1000} = 0.555$

Page No 25.6:

Question 2:

A die is thrown 100 times and outcomes are noted as given below:

Outcome:	1	2	3	4	5	6
Frequency:	21	9	14	23	18	15

If a die is thrown at random, find the probability of getting a/an.
(i) 3
(ii) 5
(iii) 4
(iv) Even number
(v) Odd number
(vi) Number less than 3.

Answer:

Total number of trials = 100

Number of times "1" comes up = 21
Number of times "2" comes up = 9
Number of times "3" comes up = 14
Number of times "4" comes up = 23
Number of times "5" comes up = 18
Number of times "6" comes up = 15

(i) Probability of getting 3

= $\frac{frequency of 3}{Total no . of trails} = \frac{14}{100} = 0.14$

(ii) Probability of getting 5

= $\frac{frequency of 5}{Total no . of trails} = \frac{18}{100} = 0.18$

(iii) Probability of getting 4

= $\frac{frequency of 4}{Total no . of trails} = \frac{23}{100} = 0.23$

(iv) Frequency of getting an even no. = Frequency of 2 + Frequency of 4 + Frequency of 6 = 9+ 23 + 15 = 47

Probability of getting an even no. = $\frac{frequency of even no .}{Total no . of trails} = \frac{47}{100} = 0.47$

(v) Frequency of getting an odd no. = Frequency of 1 + Frequency of 3 + Frequency of 5 = 21+ 14 + 18 = 53

Probability of getting an odd no. = $\frac{frequency of odd no .}{Total no . of trails} = \frac{53}{100} = 0.53$

(vi) Frequency of getting a no. less than 3 = Frequency of 1 + Frequency of 2= 21 + 9 = 30

Probability of getting a no. less than 3

= $\frac{frequency of no . less than 3}{Total no . of trails} = \frac{30}{100} = 0.30$

Page No 25.6:

Question 3:

A box contains two pair of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that I will make a pair?

Answer:

No. of socks in the box = 4
Let B and W denote black and white socks respectively.
Then we have:
S = {B,B,W,W}

If a white sock is picked out, then the total no. of socks left in the box = 3
No. of white socks left = 2-1 =1

Probability of getting a white sock

= $\frac{Number of white socks left in the box}{Total no . of socks left in the box} = \frac{1}{3}$

Page No 25.6:

Question 4:

Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:

Outcome:	Two heads (HH)	One head (HT or TH)	No head (TT)
Frequency:	105	275	120

If same pair of coins is tossed at random, find the probability of getting (i) Two heads (ii) One head (iii) No head.

Answer:

Number of trials = 500
Number of outcomes of two heads (HH) = 105
Number of outcomes of one head (HT or TH) = 275
Number of outcomes of no head (TT) = 120

(i) Probability of getting two heads = $\frac{frequency of getting 2 heads}{Total number of trails} = \frac{105}{500} = \frac{21}{100}$

(ii) Probability of getting one head = $\frac{frequency of getting 1 head}{Total number of trails} = \frac{255}{500} = \frac{11}{20}$

(iii) Probability of getting no head = $\frac{frequency of getting no head}{Total number of trails} = \frac{120}{500} = \frac{6}{25}$

Page No 25.7:

Question 1:

An unbiased coin is tossed once, the probability of getting head is

(a) $\frac{1}{2}$ (b) 1 (c) $\frac{1}{3}$ (d) $\frac{1}{4}$

Answer:

Tossing a coin, either we get a head (H) or a tail (T). So, the probability of getting a head is $\frac{1}{2}$ .
Hence, the correct option is (a).

Page No 25.7:

Question 2:

There are 10 cards numbered from 1 to 10. A card is drawn randomly. The probability of getting
an even numbered card is

(a) $\frac{1}{10}$ (b) $\frac{1}{5}$ (c) $\frac{1}{2}$ (d) $\frac{2}{5}$

Answer:

The number on the cards are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
The even numbers on the cards are 2, 4, 6, 8, 10.
∴ Probability of getting an even numbered card = $\frac{Number of even numbered card}{Number of cards with numbers from 1 to 10}$

$= \frac{5}{10} = \frac{1}{2}$
Hence, the correct option is (c).

Page No 25.7:

Question 3:

A dice is rolled. The probability of getting an even prime is

(a) $\frac{1}{6}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{5}{6}$

Answer:

The possible numbers on a dice are 1, 2, 3, 4, 5, 6.
There is only one even prime number which is 2.
∴ Probability of getting an even prime = $\frac{Number of even prime numbers}{Number of all possible outcomes on the dice}$ $= \frac{1}{6}$

Hence, the correct option is (a).

Page No 25.7:

Question 4:

There are 100 cards numbered from 1 to 100 in a box. If a card is drawn from the box and
the probability of an event is $\frac{1}{2}$ , then the number of favourable cases to the event is

(a) 20 (b) 25 (c) 40 (d) 50

Answer:

Here, $\frac{50}{100} = \frac{1}{2}$ .
So, if the the probability of an event is $\frac{1}{2}$ , then the number of favourable cases has to be 50.
Hence, the correct option is (d).

Page No 25.7:

Question 5:

When a dice is thrown, the total number of possible outcomes is

(a) 6 (b) 1 (c) 3 (d) 4

Answer:

The number on the faces of a dice are 1, 2, 3, 4, 5, and 6.
∴ Number of possible outcomes = 6
Hence, the correct option is (a).

Page No 25.7:

Question 6:

There are 10 marbles in a box which are marked with the distinct numbers from 1 to 10.
A marble is drawn randomly. The probability of getting prime numbered marble is

(a) $\frac{1}{2}$ (b) $\frac{2}{5}$ (c) $\frac{9}{3}$ (d) $\frac{3}{10}$

Answer:

The numbers marked on the marbles are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Here, the prime numbers (favourable outcomes) are 2, 3, 5, and 7.
∴ Number of favourable outcomes = 4
Therefore
Probability of getting prime numbered marble = $\frac{4}{10} = \frac{2}{5}$
Hence, the correct option is (b).

Page No 25.7:

Question 7:

The probability of getting a red card from a well shuffled pack of cards is

(a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (c) $\frac{3}{4}$ (d) $\frac{1}{3}$

Answer:

There are 52 cards in a standard deck. There are four different suits Diamonds (red), Clubs (black)
, Hearts (red), and Spades (black) each containing 13 cards.
∴ Number of red cards (favourable outcomes) = 13 + 13 = 26
Therefore
Probability of getting a red card = $\frac{26}{52} = \frac{1}{2}$
Hence, the correct option is (b).

Page No 25.7:

Question 8:

A coin is tossed 100 times and head is obtained 59 times. The probability of getting a tail is

(a) $\frac{59}{100}$ (b) $\frac{41}{100}$ (c) $\frac{29}{100}$ (d) $\frac{43}{100}$

Answer:

Number of all possible outcomes = 100
Number of head obtained = 59
Number of tail obtained (favourable outcomes) = 100 − 59 = 41
Therefore
Probability of getting a tail = $\frac{41}{100}$
Hence, the correct option is (b).

Page No 25.7:

Question 9:

A dice is tossed 80 times and number 5 is obtained 14 times. The probability of not getting the number 5 is

(a) $\frac{7}{40}$ (b) $\frac{7}{80}$ (c) $\frac{33}{40}$ (d) None of these

Answer:

Probability of getting 5 = $\frac{14}{80} = \frac{7}{40}$
Therefore
Probability of not getting 5 = $1 - \frac{7}{40} = \frac{33}{40}$
Hence, the correct option is (c).

Page No 25.7:

Question 10:

A bag contains 4 green balls, 4 red balls and 2 blue balls. If a ball is drawn from the bag, the
probability of getting neither green nor red ball is

(a) $\frac{2}{5}$ (b) $\frac{1}{2}$ (c) $\frac{4}{5}$ (d) $\frac{1}{5}$

Answer:

The probability of getting neither green nor red ball is equal to the probability of getting blue balls.
Number of blue balls = 2
Total number of balls = 4 + 4 + 2 = 10
Therefore
Probability of getting neither green nor red ball = $\frac{2}{10} = \frac{1}{5}$
Hence, the correct option is (d).

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