You have a bag containing 20 apples, 10 oranges, and an unknown number of pears. If the probability that you select 2 apples and 2 oranges is equal to the probability that you select 1 apple, 1 orange, and 2 pears, then what is the number of pears originally in the bag?

Using $n$ as the number of pears, I found the probability of selecting 2 apples and 2 oranges to be: $$\frac{\dbinom{20}{2}*\dbinom{10}{2}}{\dbinom{30+n}{4}}$$

seeing as the number of ways to choose two apples and 2 oranges (order shouldn't matter) would be given by $\dbinom{20}{2}*\dbinom{10}{2}$. I put this over the total number of possibilities, which was found by choosing $4$ fruits from a total of $30+n$ fruits.

As the problem stated, this equaled the probability of choosing 1 apple, 1 orange, and 2 pears, which would be: $$\frac{\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}}{\dbinom{30+n}{4}}$$

And thus: $$\frac{\dbinom{20}{2}*\dbinom{10}{2}}{\dbinom{30+n}{4}}=\frac{\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}}{\dbinom{30+n}{4}}$$ $$\Rightarrow\dbinom{20}{2}*\dbinom{10}{2}=\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}$$ $$\Rightarrow\dbinom{n}{2}=42.75$$

which doesn't seem to be correct. Can anyone tell me where I went wrong? I believe it is most likely a conceptual mistake; is this not how you calculate the respective probabilities?

  • 1
    $\begingroup$ it looks like your logic is correct. You're sure those are the correct numbers in question? $\endgroup$ – Kamster Feb 10 '16 at 3:39

It is good that you were alert. I believe everything you are doing is correct. If we continue, then $$\binom{n}{2} = \frac{\binom{20}{2}\binom{10}{2}}{20(10)} = \frac{171}{4}.$$ I think it is ok to have a decimal number here.

This gives \begin{align*} \frac{n!}{2!(n-2)!} &= \frac{171}{4}\\ \implies \frac{n!}{(n-2)!} &= \frac{171}{2}\\ n(n-1) &= \frac{171}{2}\\ n^2-n+\frac{1}{4} &= \frac{171}{2}+\frac{1}{4}\\ \left(n-\frac{1}{2}\right)^2 &= \frac{343}{4}\\ \implies n &= \sqrt{\frac{343}{4}}+\frac{1}{2}\\ n&=9.76013 \end{align*} which I feel is an appropriate approach. The numbers are usually chosen so that when you complete the square, you will end up with a whole number $n$, but I think there was an error.

  • $\begingroup$ Okay, thank you for your help! It was the last question on a math contest, so I may have misread the amount of apples and oranges $\endgroup$ – Junlin Yi Feb 10 '16 at 21:45
  • $\begingroup$ @JunlinYi No problem. If it was for a contest problem, then you were probably expected to recognize $n$ immediately from $\binom{n}{2} = 300$ for example, and $n$ would be $25$. Alternatively, when/if you reach $n(n-1) = 600$, then recognize the consecutive pair as $25\cdot 24$ which imlies $n = 25$. $\endgroup$ – Em. Feb 10 '16 at 22:39

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