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?