committee of 4 to be selected from 5 boys and 4 girls. What is probability that committee will be a) all girls b) more boys than girls? ?

a)

No. of combinations to form the committee without restriction
= ₉C₄
= 126

No. of combinations that the committee has all girls, i.e. (4 girls)
= ₄C₄
= 1

The required probability
= 1/126

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b)
No. of combinations that the committee has more boys than girls
= No. of combination that the committee has (4 boys) or (3 boys 1 girl)
= ₅C₄ + ₅C₃×₄C₁
= 5 + 40
= 45

The required probability
= 45/126
= 5/14

Total ways of choosing 4 from 9 is 9C4 = 126

a) If the committee is comprised of all girls, we have 4 girls and no boys.

so, 4C4 x 5C0 = 1....where choosing from girls and boys is independent.

Hence, P(all girls) = 1/126

b) More boys than girls means either 4 boys and 0 girls, or 3 boys and 1 girl

i.e. 5C4 x 4C0 or 5C3 x 4C1

=> (5 x 1) + (10 x 4)

so, 45

Hence, P(more boys than girls) = 45/126 = 5/14

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