Question

2.Exam-like questioin A class consists of 12 girls and 16 boys. Jane is one of the girls and Jonatharn is one of the boys. (i) In how many ways can we make a group of 2 girls and 3 boys? (ii) What is the probability that Jane is not in that group and Jonathan is? (iii) What is the probability that Jonathan is and Jane is not in a group of 5 students selected randomly out of the 28? (iv) What is the probability that there are at least 2 boys and at least 1 girl in a randomly selected group of 5 students?

Answer 1

(i) 2 girls can be selected from 12 girls in 12C2 ways and 3 boys can be selected from 16 boys in 16C3 ways.

Thus, total no. of ways the group can be made = 12C2 * 16C3

= 36960

(ii) No. of groups in which Jane is not in the group and Jonathan is = 11C2 * 15C2 = 5775

Thus, required probability = 5775/36960

= 5/32

(iii) No. of ways in which 5 student can be selected out of the 28 = 28C5

No. of groups in which Jonathan is and Jane is not = 26C4

Thus, required probability = 26C4/28C5

= 0.152

(iv) Such groups can have (2 boys, 3 girls), (3 boys, 2 girls), (4 boys, 1 girl)

Thus, total no. of such groups = (16C2* 12C3) + (16C3*12C2) + (16C4*12C1) = 85200

So the required probability = 85200/28C5

= 710/819 = 0.8669

Note, nCr = $\binom{n}{r}$