In a business school at ABC University, 1/6 of the enrollment are freshmen, 3/10 are sophomores, 1/3 are juniors and 1/5 are seniors. They compete with several other schools within their region in an annual financial case competition.
Three students were randomly selected to form a team for this year’s competition.
Calculate the probability that the juniors would make up the majority of the team.
I thought that this probability is equal to the probability that exactly 2 out 3 selected are juniors plus the probability that all 3 are juniors. Therefore, I computed the following:
$$P(exactly~2~juniors) = \frac13 \times \frac13 \times \frac23 = \frac2{27}$$
$$P(3~juniors) = (\frac13)^3$$
Therefore, $$P(juniors~make~up~the~majority)= \frac1{27} + \frac2{27} = \frac3{27}$$
My answer is wrong as I have only partially computed the probability. I would appreciate some explanation as what I am missing as I am having trouble figuring it out. Thank you
Best Answer
Your formula $\frac{1}{3}\cdot\frac{1}{3}\cdot \frac{2}{3}$ to pick the two juniors is the probability of first selecting two juniors, and then the non-junior. You can also first pick a junior, then a non-junior, and then a junior again, the probability of which is: $\frac{1}{3}\cdot\frac{2}{3}\cdot \frac{1}{3}$. And, of course, you can first pick the non-junior, and then the two juniors, with a probability of $\frac{2}{3}\cdot\frac{1}{3}\cdot \frac{1}{3}$.
All these give you 2 juniors and 1 non-junior, so the probability of picking exactly 2 juniors is:
$\frac{1}{3}\cdot\frac{1}{3}\cdot \frac{2}{3}+\frac{1}{3}\cdot\frac{2}{3}\cdot \frac{1}{3}+\frac{2}{3}\cdot\frac{1}{3}\cdot \frac{1}{3} = \frac{6}{27}=\frac{2}{9}$