Q. 195.0( 1 Vote )

# A committee of 4

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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