Question

One day, eleven babies are born at a hospital. Assuming each baby has an equal chance of being a boy or girl, what is the probability that at most nine of the eleven babies are girls?
A.
23/256

B.
191/256

C.
509/512

D.
397/2048

Answer 1

You only need to consider the situations where 10 or 11 of the babies are girls, then subtract those probabilities from 1.  This will give probability that any other number up to 9 of the babies are girls.

Use binomial theorem.
$P(x=k) = (nCk) p^k (1-p)^{n-k}$

n = 11
k = 10,11
p = 1/2

$P(x=10) = 11 (\frac{1}{2})^{11} = \frac{11}{2048} \\ \\ P(x=11) = 1(\frac{1}{2})^{11} = \frac{1}{2048} \\ \\ P(x \leq 9) = 1 - \frac{11}{2048} - \frac{1}{2048} \\ \\ P(x \leq 9)=\frac{2036}{2048} = \frac{509}{512}$