Probability
Conditional Probability and Bayes' Theorem
Grade 12
Question:
<p>A doctor is called to see a sick child. The doctor knows (prior to the visit) that 90% of the sick children in that neighbourhood are sick with the flue, denoted by F, while 10% are sick with the measles, denoted by M. The probability of having a rash for a child sick with the measles is 0.95. However, occasionally children with the flue also develop a rash with conditional probability 0.08. Upon examination the child, the doctor finds a rash, then the probability that the child has the measles, is</p>
<p>(a) \(\frac{89}{167}\)</p>
<p>(b) \(\frac{91}{167}\)</p>
<p>(c) \(\frac{93}{167}\)</p>
<p>(d) \(\frac{95}{167}\)</p>
Step-by-Step Solution
Key Concept: Apply Bayes' theorem to find the posterior probability of measles given that a rash is observed. The denominator includes both pathways that can produce a rash.
<p><strong>Step 1:</strong> Given: $P(F) = 0.90$, $P(M) = 0.10$, $P\left(\frac{R}{M}\right) = 0.95$, $P\left(\frac{R}{F}\right) = 0.08$</p><p><strong>Step 2:</strong> Using Bayes' theorem: $P\left(\frac{M}{R}\right) = \frac{P(M) \cdot P\left(\frac{R}{M}\right)}{P(M) \cdot P\left(\frac{R}{M}\right) + P(F) \cdot P\left(\frac{R}{F}\right)}$</p><p><strong>Step 3:</strong> $P\left(\frac{M}{R}\right) = \frac{0.10 \times 0.95}{0.10 \times 0.95 + 0.90 \times 0.08} = \frac{0.095}{0.095 + 0.072} = \frac{0.095}{0.167} = \frac{95}{167}$</p><p>∴ Answer is (d).</p>
Correct Answer: D