Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gender of any other children. find the probability of exactly eight boys in ten births. round the answer to the nearest thousandth.

Let X be the number of boys in n selected births. Let p be the probability of getting baby boy on selected birth.

Here n=10. Also the male and female births are equally likely it means chance of baby boy or girl is 1/2

P(Boy) = P(girl) =0.5

p =0.5

From given information we have n =10 fixed number of trials, p is probability of success which is constant for each trial . And each trial is independent of each other.

So X follows Binomial distribution with n=10 and p=0.5

The probability function of Binomial distribution for k number of success, x=k is given as

P(X=k) =

We have to find probability of getting 8 boys in n=10 births

P(X=8) =

= 45 * 0.0039 * 0.25

P(X = 8) = 0.0438

The probability of getting exactly 8 boys in selected 10 births is 0.044

the correct answer is 828.07

-7

step-by-step explanation:

x^2 + 18x + 74 = (x + 9)^2 - 81 + 74

= (x + 9)^2 - 7.

so q = -7.