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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.

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  • $\begingroup$ $$\frac{\binom{5}{2} \times \binom{7}{2}}{\binom{15}{4}}.$$ Denominator is number of ways of selecting $(4)$ pieces of fruit out of $(15)$. Under the constraints, exactly $(2)$ of the selected pieces of fruit must be oranges. $\endgroup$
    – user2661923
    Commented Jul 14, 2021 at 3:06
  • $\begingroup$ Why do i get a different answer if I do (5C2)(3C0)/15C4 $\endgroup$
    – Joe
    Commented Jul 14, 2021 at 3:12
  • $\begingroup$ Your $\binom{3}{0}$ factor can (harmlessly) be added as a 3rd factor in the numerator. However, you can not omit the $\binom{7}{2}$ factor, which reflects selecting $(2)$ oranges out of $(7)$. To see the relevance of this factor, ask yourself how the numerator would be affected if there were $(100)$ oranges, instead of only $(7)$. Certainly, the denominator would also be affected, however, the two changes (to the numerator and to the denominator) would not cancel each other out. $\endgroup$
    – user2661923
    Commented Jul 14, 2021 at 3:16

2 Answers 2

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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...

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

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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 $15.384615\%$.

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