# Thread: Probability Help - Negative Binomial

1. ## Probability Help - Negative Binomial

Hi

I know this problem is negative binomial, but I'm having problems on trying to set it up. Any help would be greatly appreciated.

A family decides to have children until they have three boys. Assuming that the probability of having a girl is 0.5 per birth and that the gender of each child is not influenced by the other children in the family, find the probability mass function of the random variable X = number of children in the family. Next solve the same problem where instead of having children until they have three boys the family stops when they have the third child of the same gender.

2. Originally Posted by shogunhd
Hi

I know this problem is negative binomial, but I'm having problems on trying to set it up. Any help would be greatly appreciated.

A family decides to have children until they have three boys. Assuming that the probability of having a girl is 0.5 per birth and that the gender of each child is not influenced by the other children in the family, find the probability mass function of the random variable X = number of children in the family. Next solve the same problem where instead of having children until they have three boys the family stops when they have the third child of the same gender.
Part I is simple. Using the notation given in

Negative binomial distribution - Wikipedia, the free encyclopedia,

just substitute r = 3 and p = 1/2: $\displaystyle \, \Pr(X = k) = f(k) = {k + 2 \choose k} \left( \frac{1}{2} \right)^{3 + k}$.

For Part II, let Y be the random variable number of children in family. Since the family stops when they have either 3 boys or 3 girls, Y can only have the values 3, 4 or 5:

$\displaystyle \Pr(Y = 3) = 2 \, \left( \frac{1}{2} \right)^3 = \frac{1}{4}$.

(This corresponds to BBB or GGG)

$\displaystyle \Pr(Y = 4) = (2) (3) \left( \frac{1}{2} \right)^4 = \frac{3}{8}$.

(This corresponds to BGGG, GBGG, GGBG or GBBB, BGBB, BBGB)

$\displaystyle \Pr(Y = 5) = 1 - \frac{1}{4} - \frac{3}{8} = \frac{3}{8}$.

(This corresponds to BBGGG, etc ...... or GGBBB, etc .......)