A book club sends $6$ paperback and $2$ hardback books to Mrs. Hunt. She chooses $4$ of these books at random to take with her on holiday. The random variable $X$ represents the number of paperback books she chooses.
Show that the probability that she chooses exactly $2$ paperback books is $\frac{3}{14}$.
Here's what I did:
$$ \frac{6}{8} \cdot \frac{5}{7} \cdot \frac{2}{6} \cdot \frac{1}{5}$$
Apparently I need to multiply by $4C2$.
Can someone please explain why this is $4C2$ and not $6C2$? Given that the total paperbacks are $6$, and she'll choose two from those six?
I'm a little confused here.
Best Answer
The $\binom{4}{2}$ in correcting your attempt is not in reference to which two of the six paperback books were taken, but rather in reference to which positions in the listings of the four books being taken were occupied by paperback since otherwise your answer was phrased in a way as though order mattered. Without the correcting factor, your answer of $\frac{6}{8}\times\frac{5}{7}\times\frac{2}{6}\times\frac{1}{5}$ was the probability that taking four books in sequence that very explicitly the first and second books were both paperback and the third and fourth books were both hardback. It failed to account for scenarios such as the first book being hardback, the second and third both being paperback, and the fourth being hardback, etc...
Remember that when phrasing your answer it often does not matter whether you treat order as relevant or irrelevant so long as you are consistent about it. If your numerator treats order as irrelevant your denominator should also treat order as irrelevant. If your numerator treats order as relevant your denominator should also treat order as relevant, etc...
If you choose to phrase this instead where order is irrelevant, then this is a straightforward application of hypergeometric probability:
$$\frac{\binom{6}{2}\binom{2}{2}}{\binom{8}{4}}$$
which you can see is equal to the original answer after the correcting factor was applied.