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A box has $m$ black and $n$ white balls. A ball is drawn at random and put back with $k$ additional balls of the same colour as that of the drawn ball. Now a ball is drawn again. Find the probability that it is a white ball.

I got the answer $\frac{n}{m+n}$.

I just want to understand why it's not dependent on $k$.

jgon
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Maths_Rocks
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  • This problem with different numbers was posted a few minutes ago and has already some answers. It is important that you attempt to solve this though. From your tag I assume you already know that is solved using conditional probability. – Kal S. Dec 19 '18 at 13:49
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    Possible duplicate of probability - getting a red ball in thenext draw is – Kal S. Dec 19 '18 at 13:50
  • Does the bucket now contain $n+m+K$ balls in total? – Robert Frost Dec 19 '18 at 16:53
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    Strongly disagree with those voting to close as duplicates. The OP here knows how to do the problem, and is instead asking about why there's no dependence on $k$, which isn't addressed by any of the answers in the supposed duplicate. – jgon Dec 19 '18 at 16:58
  • I edited your question for formatting. You were inconsistent with the capitalization of $k$, so I chose lower case $k$ to make it consistent. Feel free to change that if you intended $k$ to be capital. – jgon Dec 19 '18 at 17:02
  • Thanx jgon !!!! – Maths_Rocks Dec 19 '18 at 17:06
  • If the ball you drew was white then adding $k$ white balls increases probability that the next ball is white. But if the ball you drew was black then adding $k$ black balls will decrease the probability. Presumably the likelihood that the probability will increasing matches the probability will decrease and they cancel out. Will have to do the actual calculations to see why it "cancels out" – fleablood Dec 20 '18 at 00:48

1 Answers1

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Look at this probability from the other side. Consider two disjoint events. Event $B_1$ means that the second ball was in the urn from the very beginning. Event $B_2$ means that the second ball is taken out of those balls that were added to the urn. Event $A$ means that the second ball is white. Consider conditional probabilities $\mathbb P(A\mid B_1)$ and $\mathbb P(A\mid B_2)$: $$\mathbb P(A\mid B_1) = \frac{n}{n+m}$$ since no ball added is taken into account when $B_1$ happens. And also $$\mathbb P(A\mid B_2) = \frac{n}{n+m}.$$ This is the probability of the second ball to be white if it is selected from $k$ balls added. In other words, this is the probability that $k$ white balls are added into the urn. This is exactly the probability that the first ball was white.

Since conditional probabilities of $A$ under both $B_1$ and $B_2$ are the same, the probabilities of $B_1$ and $B_2$ are irrelevant and $$\mathbb P(A) = \mathbb P(B_1)\frac{n}{n+m}+\mathbb P(B_2)\frac{n}{n+m}=\frac{n}{n+m}.$$ Note that only $\mathbb P(B_1)$ and $\mathbb P(B_2)$ depend on $k$.

NCh
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