0
$\begingroup$

Three students are chosen at random from a class of ten girls and twelve boys. What is the probability that at least one is a boy given at least one is a girl.

Would this be an independent or dependent event? I believe is dependent since they are not replacing their picks thus the pool gets smaller.

If that is the case then would it be $P(A\cap B)$= P(A) * P(B|A)? I'm thinking probability of ate least one boy is P(A)=$\frac{6}{11}$, and probability of at least one girl is P(B)= $\frac{5}{11}$.

2 Answers 2

1

Let $X$ be the number of boys chosen among the three. You have $X\sim\text{Hypergeometric}(22,12,3)$. Then:

$$P(\text{at least one girl})=P(X=0)+P(X=1)+P(X=2)=\frac{6}{77}+\frac{27}{77}+\frac{33}{77}=\frac{66}{77}$$

$$P(\text{at least one boy and at least one girl})=P(X=1)+P(X=2)=\frac{60}{77}$$

$$P(\text{at least one boy}|\text{at least one girl})=\frac{P(\text{at least one girl and at least one boy})}{(\text{at least one girl})}=\frac{10}{11}$$

  • 0
    So essentially we just find the probability of getting girls and just subtract it by one to get the probability of at least one boy? – 2017-02-02
  • 0
    I think the first part is not correct.. – 2017-02-02
  • 0
    Coudn't fix it, so I deleted it completely. – 2017-02-02
0

The total number of selections in which at least one child is a boy given that at least one child is a girl is \begin{equation*} \binom{10}{1}\binom{12}{2} + \binom{10}{2}\binom{12}{1} = 660 + 540 = 1200 . \end{equation*} The total number of selections in which at least one child is a girl is \begin{equation*} \binom{22}{3} - \binom{10}{3} = 1540 + 120 = 1420 . \end{equation*} So, the probability that a child selected is a boy given that one of the children selected is a girl is \begin{equation*} \frac{1200}{1420} = \frac{60}{71} . \end{equation*}