Conditionated probability (Bayes)

Probability Level 3

There are 3 urns: $U_1, U_2$ and $U_3$ .

$U_1$ contains: 3 white balls, 2 red balls, 1 black ball.
$U_2$ contains: 2 white balls, 1 red ball, 3 black balls.
$U_3$ contains: 1 white ball, 3 red balls, 2 black balls.

The following process is performed:

A coin is thrown. If the coin flips head up, the urn $U_1$ is chosen and a ball of $U_1$ is drawn.
If the coin flips tail up when the coin is thrown, the coin is thrown again and this time if it is head up, the urn $U_2$ is chosen and a ball is taken from the urn of $U_2$ . If this time the tail comes out, the urn $U_3$ is chosen and a ball is extracted from the urn $U_3$ .

After carrying out the process, it is known that the extracted ball is white. What is the probability that the ball has been selected from the urn $U_2$ ?

This probability can be written as $\frac{a}{b}$ with $a, b$ co-prime positive integers. Enter $10a + b$ .

The answer is 29.

1 solution

Jc 506881
Jan 18, 2018

Another way to describe the ball-selection process is as follows: Preform two coin flips. If the result is HH or HT, select from urn 1. If TH, select from urn 2 and if TT, select from urn 3. There are 9 ways in which a white ball is selected, two of which are from urn 2.

HH: WWW rrb

HT: WWW rrb

TH: WW rbbb <---- urn 2

TT: W rrrb

So, given that a white ball is pulled, there's a 2/9 chance it was from urn 2.

Alternatively, we can approach this problem via conditional probabilities. Let A be the event that a ball is pulled from urn 2 and let B be the event that the ball pulled is white. We want $P(A|B)$ , and we can compute it using Bayes' theorem: $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ . $P(B|A)$ is the probability of pulling a white ball from urn 2, or $\frac{1}{3}$ . $P(A) = \frac{1}{4}$ and $P(B) = \frac{9}{24} = \frac{3}{8}$ . So, $P(A|B) = \frac{2}{9}$ .

Conditionated probability (Bayes)

The answer is 29.

1 solution

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