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

$$p=P(Boy) = 0.5 \\$$

$$q=1-p=1-P(boy)=P(Girl) = 0.5$$

Probability for at least 6 boys:

$$P(X≥6) = P(X=6) + P(X=7) + P(X+8)$$

$$+ P(X=9) + P(X=10) + P(X=11)$$

$$=\dbinom{11}{6}(0.5)^6(0.5)^{11-6}+\dbinom{11}{7}(0.5)^7(0.5)^{11-7}$$

$$+\dbinom{11}{8}(0.5)^8(0.5)^{11-8}+\dbinom{11}{9}(0.5)^9(0.5)^{11-9}$$

$$+\dbinom{11}{10}(0.5)^{10}(0.5)^{11-10}+\dbinom{11}{11}(0.5)^{11}(0.5)^{11-11}$$

$$=(462+330+165+55+11+1)(0.5)^{11}$$

$$=1024(0.5)^{11}=0.5$$

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