RD Sharma Solutions Class 7 Data Handling Iv Probability Exercise 25.1

RD Sharma Solutions Class 7 Chapter 25 Exercise 25.1

RD Sharma Class 7 Solutions Chapter 25 Ex 25.1 PDF Free Download

Exercise 25.1

Q 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?

SOLUTION:

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 = \(\frac{No.\;of\;heads}{Total\;No.\;of\;trails}\) = \(\frac{445}{1000}\)=0.445

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

Q 2.A die is thrown 100 times and outcomes are noted as given below:

Capture

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.

SOLUTION:

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. = \(\frac{Frequency\;of\;even\;number}{Total\;No.\;of\;trails}\) = \(\frac{47}{100}\) = 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. = \(\frac{Frequency\;of\;odd\;number}{Total\;No.\;of\;trails}\) = \(\frac{53}{100}\) = 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 = \(\frac{Frequency\;of\;number\;less\;than\;3}{Total\;No.\;of\;trails}\) = \(\frac{30}{100}\) = 0.30

Q 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?

SOLUTION:

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 = \(\frac{no.\;of\;white\;socks\;left\;in\;the\;box}{total\;no.\;of\;socks\;left\;in\;the\;box}\) = \(\frac{1}{3}\)

Q 4.Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:

If same pair of coins is tossed at random, find the probability of getting:

(i) Two heads (ii) One head (iii) No head.

SOLUTION:

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 = \(\frac{Frequency\;of\;getting\;2\;heads}{Total\;No.\;of\;trails}\) = \(\frac{105}{500}\) = \(\frac{21}{100}\)

(ii) Probability of getting one head = \(\frac{Frequency\;of\;getting\;1\;heads}{Total\;No.\;of\;trails}\) = \(\frac{275}{500}\) = \(\frac{11}{20}\)

(iii) Probability of getting no head = \(\frac{Frequency\;of\;getting\;no\;heads}{Total\;No.\;of\;trails}\) = \(\frac{120}{500}\) = \(\frac{6}{25}\)


Practise This Question

The number system which starts with a predecessor of 1 is known as _____.