Answer to Question #273378 in Statistics and Probability for Kimutai

Question

In a family of 11 children, what is the probability that there will be more boys than girls?

Answer 1

n=11

To get more boys than girls means at least 6 boys should be there for this condition

$P(Boy) = P(X) = \frac{1}{2} = 0.5 \\ P(Girl) = P(Y) = \frac{1}{2} = 0.5$

Probability for at least 6 boys:

$P(X≥6) = P(X=6) + P(X=7) + P(X+8) + P(X=9) + P(X=10) + P(X=11) \\ P(X=6) = C^{11}_6 (0.5)^6(1-0.5)^{11-6} = 0.2255859 \\ P(X=7) = C^{11}_7 (0.5)^7(1-0.5)^{11-7} = 0.1611328 \\ P(X=8) = C^{11}_8 (0.5)^8(1-0.5)^{11-8} = 0.0805664 \\ P(X=9) = C^{11}_9 (0.5)^9(1-0.5)^{11-9} = 0.0268555 \\ P(X=10) = C^{11}_{10} (0.5)^{10}(1-0.5)^{11-10} = 0.0053711 \\ P(X=11) = C^{11}_{11} (0.5)^{11}(1-0.5)^{11-11} = 0.0004883 \\ P(X≥6) = 0.50$

There is 50% probability that there is more boys than girls.