A box of 12 items has 4 defective ones. The items are tested one by one in a random order, without replacement. What is the probability that the first defective item is found on the third test?

\frac{4}{27}

Correct answerYour answer

\frac{28}{165}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{14}{33}

Correct answerYour answer

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CorrectNot quiteAnswer B

Correct answer: B

The first two items must be good and the third defective: $\frac{8}{12} \times \frac{7}{11} \times \frac{4}{10}$ .
$= \frac{224}{1320} = \frac{28}{165}$ .

Worth noting

$\frac{4}{27} = \left(\frac{2}{3}\right)^{2} \times \frac{1}{3}$ keeps the box unchanged between tests, but a tested item is not put back.

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

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