I am reading a problem about probabilities with dependent events.
There are $3$ people selected from $5$ freshmen, $2$ sophomores and
$4$ juniors. What is the probability that the first $2$ people
selected are freshmen and the third a junior?
I thought that this is solvable by using combinations.
I.e.
There are $\binom{11}{3}$ ways to select $3$ out of $11$ students (I hope that the notation is for combinations and not permutations as I get these 2 notations mixed).
This is $$\frac{11!}{3!\cdot 8!} = 165$$ ways
To choose $2$ freshman from $5$ is: $$\frac{5!}{2!\cdot 3!}$$
To choose $1$ junior from $4$ is: $$\frac{4!}{1!\cdot 3!}$$
So the probability we are looking for is:
$$\frac{\frac{5!}{2!\cdot 3!} \cdot \frac{4!}{1!\cdot 3!}}{165} = \frac{40}{165}=\frac{8}{33}$$
but the answer is wrong.
Is it not possible to use the combinations formula for calculating such a probability with $3$ dependent events?
Best Answer
Your answer would be correct if the question were:
That would be a question about combinations - how many ways to choose 2 freshman out of 5 and one junior out of 4, over the number of ways to choose a group of 3 students from 11.
That is subtly different from the question which was asked. The question in your post says the first two people are freshman, and the third is a junior - i.e., the order of selection matters. So you should instead use permutations.