probability with selecting fruit A box contains 3 apples, 7 oranges, and 5 bananas. 4 fruit are selected at random without replacement (Assuming that you cannot tell the difference between the fruit when you select them).
What is the probability that exactly 2 are bananas and none are apples?
I have tried the formula (aCx)(n-aCr-x)/nCr to find the probability of exactly 2 bananas and exactly 0 apples separately, then multiplied the two to get approx. 12 percent. I doubt this is correct.
 A: You cannot simply multiply the probabilities as the events are clearly not independent.

The hypergeometric pmf measures the probability for obtaining $x$ from $a$ favoured and $r-x$ from $n-a$ unfavoured when selecting $r$ from all $n$ items.  Now, if we let $y=r-x$ and $b=n-a$ then we get the probability for obtaining $x$ from $a$ in category-1 and $y$ from $b$ in category-2 when selecting $x+y$ from all $a+b$ items:
$$\dfrac{\dbinom ax\dbinom{n-a}{r-x}}{\dbinom{n}{r}}=\dfrac{\dbinom ax\dbinom{b}{y}}{\dbinom{a+b}{x+y}}$$
Extending this to three (or more) categories is simply a matter of matching the number of binomial coefficients in the numerator to the number of categories.  (ie Use three of them.)
You seek the probability for obtaining 0 from 3 apples, 2 from 5 bananas, and 2 from 7 oranges when selecting 4 from all 15 fruit.
So...

 $$\mathsf P(A=0,B=2,O=2)=\dfrac{\dbinom 30\dbinom 52\dbinom72}{\dbinom{3+5+7}{0+2+2}}$$

A: My logic is as follows. The probability of choosing 2 bananas and no apples in some way (take b,b,a,a for example) is:
\begin{align*}
(\frac{5}{15}\frac{4}{14})(\frac{7}{13}\frac{6}{12}) = \frac{840}{32760}
\end{align*}
Then since there are $\binom{4}{2}$ ways of getting two bananas and two apples, we need to multiply the probability of one possible order happening by the number of possible orders to get the total probability.
$$\frac{840}{32760}\binom{4}{2}\approx 0.15384615$$
So the probability is about $15.384615\%$.
