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?

1. minhkhoi
The probability of having, at most, 9 girls, is 0.9515

### How to get the probability?

The probability that a random baby is a girl is:
p = 0.5
And the probability that a random baby is a boy is:
q = 0.5
Then the probability that, at most, 9 out of 11 babys are girls, is given by:
1 – p(10) – p(11)
Where P(10) is the probability that 10 of the babies are girls and p(11) is the probability that the 11 babies are girls.
p(10) = C(11, 10)*(0.5)^10*(0.5)^1 = C(11, 9)*(0.5)^11
Where C(11, 10) is the combinations of 10 elements that we can make with a set of 11 elements, such that:
C(11, 10)=  11!/(11 – 10)!*10! = 11
Replacing that, we get:
P = 11*(0.5)^11 = 0.0054
p(11) = C(11, 11)*0.5^11 = 1*0.5^11 = 0.0005
Then the probability is:
P = 1 – 0.0054 – 0.0005 = 0.9515
The probability of having, at most, 9 girls, is 0.9515