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In a bag there are $15$ red and $5$ white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is?

I have attempted this question by counting all the favorable cases:

  1. Both red $(15×14)$
  2. One red one white $(15×5)$

Our case is both red. The probability is, by Baye's theorem, $\dfrac{15×14}{15×14+15×5}$. However, the answer is not $\dfrac{14}{19}$ but $\dfrac{7}{12}$.

Ѕᴀᴀᴅ
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2 Answers2

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You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$\frac{15\cdot14}{15\cdot14+2\cdot15\cdot5}=\frac7{12}.$$

Edit: This is assuming that both balls were inspected and (at least) one is found to be red.
The answer is $\frac{14}{19}$ if one ball was picked from the two chosen and it is found to be red.

Abraham Zhang
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  • Yes, the question is ambiguous as stated, which often happens in these sorts of conditional probability problems. Most of us would assume the condition is "at least one red ball was picked" but it certainly could be "a randomly selected ball from the two picked turned out to be red." – Ned Mar 17 '19 at 01:09
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In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15\cdot 5 + 5\cdot 15 = 150$. So we have $15\cdot 4 + 15\cdot 5 + 5\cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.

mlerma54
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