How do I work backwards?

Probability Level 2

Box I contains 1 red ball and 2 white balls. Box II contains 2 red balls and 1 white ball. One ball is drawn randomly from box I and transferred to box II.

Then, a ball is drawn randomly from box II and it is red. What is the probability that the transferred ball was white?

0.57 0.14 1 2/3

2 solutions

Dan Wilhelm
Jul 13, 2015

Let the event $W$ be that the transferred ball is white. Let the event $R$ be that the ball drawn is red. Then, from counting we are given that $P(W) = 2/3$ and $P(R|W) = 2/4 = 1/2$ .

The chance of drawing a red ball is dependent on whether a red or white ball was transferred. So:

$P(R) = P(W)P(R|W) + P(\neg W)P(R|\neg W) = \tfrac{2}{3} \cdot \tfrac{1}{2} + \tfrac{1}{3} \cdot \tfrac{3}{4} = \tfrac{7}{12}$ .

Applying Bayes' Rule:

$P(W|R) = \frac{P(R|W)P(W)}{P(R)} = \frac{\tfrac{1}{2} \cdot \tfrac{2}{3}}{\tfrac{7}{12}} = \frac{4}{7}$ .

Hence, $P(W|R) = 4/7$ .

I don't understand this problem. The question implies that the fact that a red ball was drawn from Box 2 is irrelevant because all it asks for is the probability that a white ball was transferred from Box 1 to 2. Drawing a red ball after the fact has no bearing on the question. Please help me out.

A Former Brilliant Member - 1 year, 8 months ago

What we draw (uniformly randomly) is directly affected by what we transfer. Drawing a R from Box 2 is more likely if we transfer a R!

Suppose we initially have 0R 1000W in Box 2.

If we then draw a R from Box 2, there is a 100% chance we transferred a R.
If we instead draw a W from Box 2, there is a (slightly less than) 50% chance we transferred a R.

So, what we draw gives us information about what we transferred.

Here is perhaps a clearer case-by-case explanation:

Box 1 starts with 1R 2W.
Box 2 starts with 2R 1W.

We draw one ball randomly from Box 1 and transfer it to Box 2. So each of these has 1/3 probability of occurring:

(A) Transfer R => Box 2 - 3R 1W <---- 3/4 chance of drawing R
(B) Transfer W => Box 2 - 2R 2W <---- 1/2 chance of drawing R
(C) Transfer W => Box 2 - 2R 2W <---- 1/2 chance of drawing R

Now, we randomly draw a R from Box 2. Since (A), (B), and (C) occur equally, the probability of transferring a W is:

$(1/2 + 1/2) / (3/4 + 1/2 + 1/2) = 1 / (7/4) = 4/7$ .

Dan Wilhelm - 1 year, 7 months ago

Got it. Thanks!

A Former Brilliant Member - 1 year, 7 months ago

Ted Schaft
Dec 12, 2017

The probability of drawing a red ball from Box I is 1/3, this will make the probability of drawing a red ball from Box II 3/4. P(RR =1/4). The probability of drawing a white ball from Box I is 2/3, this will make the probability of drawing a red ball from Box II 1/2. P(WR = 1/3).

These are the only cases that count as they have a red ball drawn from Box II. The probability of having a white ball transferred given a red ball drawn from Box II is (1/3)/(1/3 + 1/4) = 4/7

nice solution!

Laura Gao - 2 years, 10 months ago

How do I work backwards?

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