Question

Suppose that 45% of all babies born in a particular hospital are girls. If 7 babies born in the hospital are randomly selected, what is the probability that at most 2 of them are girls?Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Answer 1

Solution:

Since the sample size is fixed;

$n=7$

And the probability of picking a girl is constant for each trial, the binomial distribution is appropriate;

$\begin{gathered} P(X=r)=^nC_rP^r(1-P)^{n-r} \\ \\ P(X\leq2) \end{gathered}$

Thus;

$\Rightarrow^7C_0(0.45)^0(0.55)^7+^7C_1(0.45)^1(0.55)^6+^7C_2(0.45)^2(0.55)^5$$\Rightarrow0.0152+0.0872+0.2140=0.3164$

ANSWER: 0.32