Problem 5.12 Easy Free 2 variations

Diagnostic Test Base Rate

A disease has prevalence $2\%$ in a population. A diagnostic test for the disease has a false negative rate of $10\%$ (it misses $10\%$ of true cases) and a false positive rate of $1\%$ (it wrongly flags $1\%$ of healthy people). Given that a randomly chosen person tests positive, find the probability that they actually have the disease.

Work it out, then compare with the answer.

SolutionShow the worked solution

Let $D$ be the event that the person has the disease and $+$ the event that the test is positive. We are given

\mathbb{P}(D)=0.02,\qquad \mathbb{P}(+\mid D)=1-0.10=0.9,\qquad \mathbb{P}(+\mid D^c)=0.01 .

By the law of total probability, the probability of a positive test is

\mathbb{P}(+)=0.9\cdot 0.02+0.01\cdot 0.98=0.018+0.0098=0.0278 .

Bayes’ rule then gives

\mathbb{P}(D\mid +)=\frac{\mathbb{P}(+\mid D)\,\mathbb{P}(D)}{\mathbb{P}(+)}=\frac{0.018}{0.0278}=\frac{180}{278}=\frac{90}{139}\approx 0.647 .

In natural frequencies, take $10\,000$ people. Of these, $200$ are ill and $180$ of them test positive. The other $9800$ are healthy and $98$ of them test positive. So $180$ of the $278$ positives are genuine.

\boxed{\mathbb{P}(D\mid +)=\dfrac{90}{139}\approx 0.647}

Key idea. The test is $99\%$ specific, but the healthy group is $49$ times larger than the ill group, so false positives make up more than a third of all positives.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 5.12a

Among a large pool of job candidates, $45\%$ never used a certain prep platform, $30\%$ used it lightly, and $25\%$ used it heavily. Light users are twice as likely to receive a job offer as non-users, and heavy users are twice as likely to receive an offer as light users. Given that a candidate received an offer, find the probability that they were a heavy user of the platform.

Work it out, then compare with the answer.

Solution

Let $p$ be the offer probability for a non-user. Then a light user gets an offer with probability $2p$ and a heavy user with probability $4p$ . By Bayes’ rule,

\mathbb{P}(\text{heavy}\mid\text{offer})=\frac{4p\cdot 0.25}{p\cdot 0.45+2p\cdot 0.30+4p\cdot 0.25}=\frac{1}{0.45+0.6+1}=\frac{1}{2.05}.

The unknown $p$ cancels, so only the ratios matter.

\boxed{\dfrac{20}{41}\approx 0.488}

Variation 5.12b

A letter is delivered to one of three mailboxes, numbered $1$ , $2$ , $3$ , each chosen with equal probability. If the letter lands in mailbox $i$ , it gets read with probability $i/3$ . You check mailbox $1$ and find no letter has been read there. Find the probability that the letter was in fact delivered to mailbox $1$ .

Work it out, then compare with the answer.

Solution

Let $M_i$ be the event that the letter was delivered to box $i$ , and $N$ the event that no letter is read in box $1$ .

If the letter is in box $1$ , it goes unread with probability $\tfrac23$ . If it is in box $2$ or $3$ , nothing can be read in box $1$ , so $N$ is certain. Hence

\mathbb{P}(M_1\mid N)=\frac{\tfrac13\cdot\tfrac23}{\tfrac13\cdot\tfrac23+\tfrac13+\tfrac13}=\frac{2/9}{8/9}.

\boxed{\dfrac14}

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