The problem: Three boxes are give: A,B,C. In the box a there are two red marbles. In the box B there are two white marbles and in the box C there are one white and one red marble. One marble is randomly pulled out from one of the boxes. What is the probability that there is red marble left in the box if it's known that the marble that was first pulled out was white?

So, first of all, I am not sure if I get the problem right, I need to calculate the probability that box B is randomly chosen, because otherwise there will be not-red marble left ( in case the first one pulled out is white one). So, I calculated the probability, but I am quite sure it is incorrect as it is too easy: First I calculate the probability that the white marble is pulled out: $3/6=1/2$ as there are 3 white marbles and totally 6 marbles. Then I calculate the probability that the white marble comes from box C as that would be the only chance that in the box only red marble is left: $1/6$. Then I subtract the case when both white marble and white marble from box C is pulled out t.i. intersection: $P(A)=\frac{1}{2}+\frac{1}{6}-\frac{1}{2\cdot6}=\frac{7}{12}.$ Please correct me, if I have misunderstood the task and / or my solution is wrong.


your question is a little difficult to understand. But what I understood is that you want to find the probability of picking a box at random taking out a ball and the ball remaining is red, given the one you took out is white.

The only way in which this could happen is if you picked box $C$ and you took a white ball. The probability of that is

$$ P(\text{white out},\text{red remaining})=P_{C,white}=1/3*1/2=1/6 $$

Now, the probability of taking out a withe ball is $P(\text{white out})=1/2$ (as you computed). Then $$ P(\text{red remaining}|\text{white out})=\frac{P(\text{white out},\text{red remaining})}{P(\text{white out})}=1/3 $$

This can also be seen by noticing that when you pick a white ball, you leave a red ball if and only if you pick the box C. And this has probability 1/3.

I tried to read at solution but I don't understand your argument. In particular it seems like you are adding and multiplying probabilities that are not independent.

  • $\begingroup$ @N.F.Taussig Corrected it. Thanks :) $\endgroup$ – David Jaramillo Oct 13 '19 at 18:17

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