A die is thrown. If it shows 1, a ball is drawn from urn I (2 white, 3 black); if it shows 2 or 3, from urn II (4 white, 1 black); otherwise from urn III (3 white, 2 black). The ball drawn is white. What is the probability that it came from urn II?

\frac{1}{3}

Correct answerYour answer

\frac{8}{19}

Correct answerYour answer

\frac{4}{9}

Correct answerYour answer

\frac{4}{5}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

$P(W) = \frac{19}{30}$ , and $P(\text{II})\,P(W \mid \text{II}) = \frac{2}{6} \times \frac{4}{5} = \frac{8}{30}$ .
$P(\text{II} \mid W) = \frac{8/30}{19/30} = \frac{8}{19}$ .

Worth noting

$\frac{1}{3} = \frac{2}{6}$ is urn II's chance before the ball is seen; urn II gives white most often, so a white ball raises it.

Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.

Explanation