Answer :

Given: In a group there are 7 boys and 4 girls

Let A denote the event that exactly 2 boys will be chosen, and B the event that at least one girl will be chosen.


We require P (A | B).


When a committee of 4 student is selected exactly two boys, it means rest two selection only for girls.


So, the number of way of selection of boys = 7C2


No. of ways of selection of girls = 4C2


So, favourable outcomes = 7C2 × 4C2


Total outcomes = 11C4



Hence, P(A B) =


Now, we use the combination formula .i.e.








When at least one girl is selected in committee


There are four cases possible,


case 1:- committee selects 1 girl, 3 boys


number of ways = 4C1 × 7C3




= 4 × 35


= 140


Case 2 :- committee selects 2 girls, 2 boys


Number of ways = 4C2 × 7C2




= 6 × 21


= 126


case 3:- committee selects 3 girls, 1 boy


Number of ways = 4C3 × 7C1




= 4 × 7


= 28


Case 4 :- committee selects, 4 girls, 0 boy


number of ways = 4C4 × 7C0



= 1 [ 0! = 1]


Total outcomes = 11C4





= 330


Hence, P(B)









OR



We know that,


Sum of the probabilities = 1



C + 2C + 2C + 3C + C2 + 2C2 + 7C2 + C = 1


9C + 10C2 = 1


10C2 + 9C – 1 = 0


10C2 + 10C – C – 1


10C (C + 1) – 1(C + 1) = 0


(10C – 1)(C + 1)



Now, we have to find: E(X)


We know that, μ = E(X)



or


E(X) = ΣXP(X)


= 0 × C + 1 × 2C + 2 × 2C + 3 × 3C + 4 × C2 + 5 × 2C2 + 6 × (7C2 + C)


= 56C2 + 21C



= 0.56 + 2.1


= 2.66

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