Probability – Calculating Probability of Drawing from Box B

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I have the exercise:

Consider the following scenario. There are two boxes, Box A and Box B. Initially, Box A contains 3 red balls and 3 white balls; Box B contains 6 red balls. We swap the balls in the boxes via the following procedure: We draw a ball, denoted as ball $x_A$, uniformly at random from Box A, and draw a ball, denoted as ball $x_B$, uniformly at random from Box B. We then place ball $x_A$ into Box B and place ball $x_B$ into Box A.

After that, we select either Box A or Box B by some uniformly random procedure (like flipping a fair coin), and draw a ball from the selected box. The ball drawn was white. What was the probability that Box B was selected?

I use Bayes theorem and Arrive at 1/6

But apparently the correct answer is 11/18, how is that?

Best Answer

It is convenient to rephrase the second stage (choosing a uniformly random box and then a uniformly random ball from the box) as just choosing a uniformly random ball from the balls in both boxes.

This makes it easy to see that the probability of drawing a white ball is $\frac3{12} = \frac14$ no matter what: there are $12$ balls, and $3$ of them are white.

The probability of drawing a white ball from box B is $\frac12 \cdot \frac1{12} = \frac1{24}$. To begin with, we must end up choosing $x_A$ to be a white ball, so that box B contains any white ball to begin with. Then, in the second stage, we must choose ball $x_A$ (out of $12$ equally likely balls) in order to draw a white ball from box B.

Therefore the conditional probability is $\frac{1/24}{1/4} = \frac16$.

On the other hand, $\frac{11}{18}$ is a very closely related probability: it is the probability that we drew from box B, given that the ball we drew was red.

Here, the logic is the same. The overall probability of drawing a red ball is $\frac9{12} = \frac34$. The probability of drawing a red ball from box B is $\frac12(\frac{6}{12} + \frac5{12}) = \frac{11}{24}$, because

  • half the time, $x_A$ was red and so the red balls in box B are $6$ of the $12$ balls present;
  • half the time, $x_A$ was white and so the red balls in box B are $5$ of the $12$ balls present.

Dividing, we get $\frac{11/24}{3/4} = \frac{11}{18}$.

It seems likeliest that the problem is either misstated or misquoted and that this was the intended question: if we draw a red ball, what is the probability that we drew it from box B?

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