In a town, 60% of the people are vaccinated against flu. During a season, a vaccinated person catches flu with probability 0.1 and a person who is not vaccinated catches it with probability 0.2. A person of the town is chosen at random. Consider the following statements:
(A) The probability that the person catches flu is 0.14.
(B) The probability that the person does not catch flu is 0.84.
(C) If the person catches flu, the probability that the person was vaccinated is \(\displaystyle\frac{3}{7}\).
(D) If the person does not catch flu, the probability that the person was not vaccinated is \(\displaystyle\frac{16}{43}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
Let \(E_1\) be the event that the person is vaccinated, \(E_2\) that the person is not, and F that the person catches flu.
\(P(E_1)P(F|E_1) = 0.6\times 0.1 = 0.06\)
\(P(E_2)P(F|E_2) = 0.4\times 0.2 = 0.08\)
By the theorem of total probability:
\(P(F) = 0.06 + 0.08 = 0.14\)
\(P(F') = 1 - 0.14 = 0.86\)
By Bayes’ theorem, using \(P(F'|E_2) = 0.8\):
\( \displaystyle P(E_1|F) = \frac{0.06}{0.14} = \frac{3}{7} \)
\( \displaystyle P(E_2|F') = \frac{0.4\times 0.8}{0.86} = \frac{32}{86} = \frac{16}{43} \)
So P(flu) = 0.14, P(no flu) = 0.86, P(vaccinated | flu) = \(\frac{3}{7}\) and P(not vaccinated | no flu) = \(\frac{16}{43}\).
Checking the statements
(A) The probability that the person catches flu is 0.14. — Correct. 0.06 + 0.08 = 0.14.
(B) The probability that the person does not catch flu is 0.84. — Incorrect. It is 1 − 0.14 = 0.86.
(C) If the person catches flu, the probability that the person was vaccinated is \(\frac{3}{7}\). — Correct. \(\frac{0.06}{0.14} = \frac{3}{7}\).
(D) If the person does not catch flu, the probability that the person was not vaccinated is \(\frac{16}{43}\). — Correct. \(\frac{0.32}{0.86} = \frac{16}{43}\).