Probability of drawing from box B

Question

I have the exercise:

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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?

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I use Bayes theorem and Arrive at 1/6

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

Answer 1

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 $\frac{1/24}{1/4} = \frac16$.

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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?

Answer 2

Box A contains 3 red balls and 3 white balls; Box B contains 6 red balls.

1. Pick a ball from box A, it is red. Put it in box B. Pick a ball from box B, it is red. Put it in box A. Box A: 3 red balls + 3 white balls. Box B: 6 red balls. No change. Probability of this happening = $\frac{1}{2} \times 1 = \frac{1}{2}$

2. Pick a ball from box A, it is white. Put it in box B. Pick a ball from box B, it is red. Put it in box A. Box A: 4 red balls + 2 white balls. Box B: 5 red balls + 1 white ball. Change. Probability of this happening = $P(E1) = \frac{1}{2} \times \frac{6}{7}= \frac{6}{14}$

3. Pick a ball from box A, it is white. Put it in box B. Pick a ball from box B, it is white. Put it in box A. Box A: 3 red balls + 3 white balls. Box B: 6 red balls. No change. Probability of this happening = $P(E2) = \frac{1}{2} \times \frac{1}{7} = \frac{1}{14}$

There's only 1 scenario (out of 3) in which the number of white balls > 0 in box B. The probability that if you pick a white ball that it is box B is, cogito, low.

After the above moves have been made, you pick a ball, it is white.

$P(Box \space B|White) = P(Box \space B)\times P(White| Box \space B) \div P(White) = \frac{1}{2} \times \frac{1}{6} \div \left(\frac{2}{6} + \frac{1}{6}\right) = \frac{1}{12} \div \frac{1}{2} = \frac{1}{12} \times 2 = \frac{1}{6}$

The probability of scenario 2 occurring and you picking a white ball from box B = $\frac{6}{14} \times \frac{1}{6} = \frac{1}{14}$

Probability that the box is B given that a red ball is drawn = $P(Box \space B|Red) = P(Box \space B) \times P(Red| Box \space B) \div P(Red)$:

1. $P(E4) = \frac{1}{2} \times 1 \div \left(\frac{3}{6} + 1\right) = \frac{1}{2} \times \frac{6}{9} = \frac{1}{3}$

2. $P(E5) = \frac{1}{2} \times \frac{5}{6} \div \left(\frac{4}{6} + \frac{5}{6}\right) = \frac{5}{12} \times \frac{6}{9} = \frac{5}{18}$

$P(Box \space B|Red) = P(E4) + P(E5) = \frac{1}{3} + \frac{5}{18} = \frac{11}{18}$