RD Sharma solution class 7 chapter 25 Data Handling IV(Probability) Exercise-25.1

Exercise-25.1

Page-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 1:

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 = $\frac{\mathrm{Number} \mathrm{of} \mathrm{heads}}{\mathrm{Total} \mathrm{no}. \mathrm{of} \mathrm{trails}}=\frac{445}{1000}=0.445$

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

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 2:

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

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

(ii) Probability of getting 5

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

(iii) Probability of getting 4

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

(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. =$\frac{\mathrm{frequency} \mathrm{of} \mathrm{even} \mathrm{no}.}{\mathrm{Total} \mathrm{no}. \mathrm{of} \mathrm{trails}}=\frac{47}{100}=0.47$


(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. =$\frac{\mathrm{frequency} \mathrm{of} \mathrm{odd} \mathrm{no}.}{\mathrm{Total} \mathrm{no}. \mathrm{of} \mathrm{trails}}=\frac{53}{100}=0.53$


(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

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

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 3:

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

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

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 4:

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 =$\frac{\mathrm{frequency} \mathrm{of} \mathrm{getting} 2 \mathrm{heads}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trails}}=\frac{105}{500}=\frac{21}{100}$

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

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