# Math Help - Probability?

1. ## Probability?

The probability that any child in a certain family will have blue eyes is 1/4, and this feature is inherited independently by different children in the family. If there are five children in the family and it is known that at least one of these children has blue eyes, what is the probability that at least three of the children have blue eyes?

2. ## Re: Probability?

This is a Binomial Distribution with p = 1/4 and n = 5. What is Pr(1 <= X <= 3)?

3. ## Re: Probability?

Conditional probability $P(\text{at least }3\text{ blue eyed children} | \text{at least }1\text{ blue eyed child}) = \dfrac{P(\text{at least }3\text{ blue eyed children AND at least }1\text{ blue eyed child})}{P(\text{at least }1\text{ blue eyed child})}$

If there are at least 3 blue eyed children, then there is automatically at least 1 blue eyed child. So, the numerator becomes $P(\text{at least }3\text{ blue-eyed children})$.

The probability of at least 3 blue-eyed children is the probability of exactly 3 blue-eyed children plus the probability of exactly four blue-eyed children plus the probability of exactly 5 blue-eyed children.

The probability of at least 1 blue-eyed child is the complement of no blue-eyed children.

\begin{align*}P(\text{at least }3\text{ blue-eyed children}) & = \\ \binom{5}{3}\left(\dfrac{1}{4}\right)^3\left( \dfrac{3}{4} \right)^2 + \binom{5}{4}\left(\dfrac{1}{4}\right)^4\left( \dfrac{3}{4} \right)^1 + \binom{5}{5}\left(\dfrac{1}{4}\right)^5\left( \dfrac{3}{4} \right)^0 & = \dfrac{53}{512}\end{align*}

$P(\text{at least }1\text{ blue eyed child}) = 1 - \binom{5}{0}\left(\dfrac{1}{4}\right)^0\left( \dfrac{3}{4} \right)^5 = \dfrac{781}{1024}$

So, the conditional probability:

\begin{align*}P(\text{at least }3\text{ blue eyed children} | \text{at least }1\text{ blue eyed child}) & = \\ \dfrac{\left(\dfrac{53}{512}\right)}{\left(\dfrac{ 781}{1024}\right)} & = \dfrac{106}{781} \approx 13.57\% \end{align*}

4. ## Re: Probability?

Thank you for the solutions friends.