In class 10A there are 30 boys and 20girls. In 10B there are 40 boys a

1.	In class 10A there are 30 boys and 20girls. In 10B there are 40 boys and 30girls. One student is to be selected from each class. In how many different ways we can select a pair of students. What is the probability of both being boys. What is the probability of getting one boy and one girl. What is the probability of getting at least one boy.
Answer» i. Total no. of possible pairs = 50 × 40 = 2000 No. of pairs in which both are girls = 20 × 25 = 500 Probability of both being girls = \(\frac{500}{2000}=\frac{1}{4}\) ii. No. of pairs in which both are boys = 30 × 15 = 450 Probability of both being boys = \(\frac{450}{2000}=\frac{9}{40}\) iii. No. of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050 Probability of one being boy and one girl = \(\frac{1050}{2000}=\frac{21}{40}\) iv. No. of pairs in which atleast one is boy = 1050 + 450 = 1500 Probability in which atleast one is boy = \(\frac{1050}{2000}=\frac{3}{4}\)

In class 10A there are 30 boys and 20girls. In 10B there are 40 boys and 30girls. One student is to be selected from each class. In how many different ways we can select a pair of students. What is the probability of both being boys. What is the probability of getting one boy and one girl. What is the probability of getting at least one boy.

Answer»

i. Total no. of possible pairs = 50 × 40 = 2000

No. of pairs in which both are girls = 20 × 25 = 500

Probability of both being girls = \(\frac{500}{2000}=\frac{1}{4}\)

ii. No. of pairs in which both are boys = 30 × 15 = 450

Probability of both being boys = \(\frac{450}{2000}=\frac{9}{40}\)

iii. No. of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050

Probability of one being boy and one girl = \(\frac{1050}{2000}=\frac{21}{40}\)

iv. No. of pairs in which atleast one is boy = 1050 + 450 = 1500

Probability in which atleast one is boy = \(\frac{1050}{2000}=\frac{3}{4}\)