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I have a problem stating

"There are 13 apples and 17 oranges. What is the probability of making a group of 4 with at least two apples?"

Using my logic (left side of the following equation), I first multiply a group of 2 that consists of 2 apples, and then another group of 2, which consists of the other 28 elements.

However, this is wrong; the right way to do it is to add all possible outcomes (AAOO, AAAO and AAAA) as shown on the right side of the equation

$$\frac{\binom{13}{2}\binom{28}{2}}{\binom{30}{4}} \neq \frac{\binom{13}{4}+17\binom{13}{3}+\binom{13}{2}\binom{17}{2}}{\binom{30}{4}}$$

Why is my way incorrect? Is there a smarter way to do this?

p.s.: the right answer is $\frac{1079}{1827}$ if it helps...

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    $\begingroup$ You are over counting. If your collection has, say, exactly three apples then there is no way to tell which two of those were chosen in the pair of apple, and which was chosen as part of the rest. So you end up counting that collection $3$ times. $\endgroup$
    – lulu
    Jan 22 at 14:35
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    $\begingroup$ Side note: depending on the numbers, it can save time to work with the complement of the desired selections. Here, the complement consists of those selections with no apples plus those with exactly one. That's only two types instead of the three the official solution uses. $\endgroup$
    – lulu
    Jan 22 at 14:37

2 Answers 2

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You are counting the cases with more than three apples multiple times. If three of the apples are $a,b,c$ you count picking $a,b$ for the first pair and picking $c$ plus an orange, then you count $a,c$ as the first pair and $b$ plus an orange, then you count $b,c$ as the first pair and $a$ plus an orange. Similarly you count four apples six times.

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You can use complements. That is, if $E$ is the event that there are at least two apples, then $E^c$ is the event that there are less than two apples - one or no apple, to be specific. The probability that there is only one or no apple is $$P(E^c) = \frac{\binom{17}{3}\binom{13}{1} + \binom{17}{4}\binom{13}{0}}{\binom{30}{4}} = \frac{11220}{27405}.$$

The probability of the event $P(E)$, given the probability of the complement $P(E^c)$, is given by \begin{align*}P(E) &= 1 - P(E^c) \\ P(E) &= 1 - \frac{11220}{27405} \\ P(E) &= \frac{16185}{27405} \\ P(E) &= \frac{1079}{1827}\end{align*} which is easier to compute than the event itself which has three cases.

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