Question

Suppose that 55% of all babies born in a particular hospital are boys. If 7 babies born in the hospital are randomly selected, what is the probability that fewer than 3 of them are boys? 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, i.e. $n=7$ and that the probability of picking a boy is constant for each trial (picking a baby boy) the binomial distribution is appropriate

$$P(X=r) = \ _nC_r \ P^r(1-P)^{n-r}$$

$P(X <3)$ =

$$_7C_0*0.55^0*0.45^7+ _7C_1*0.55^1*0.45^6+ _7C_2*0.55^2*0.45^5$$

$=0.2745$

The probability of getting fewer than 3 boys is 0.2745