Probability Question 6
Question 6 - 25 January - Shift 2
$25 %$ of the population are smokers. A smoker has 27 times more chances to develop lung cancer then a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $\frac{k}{10}$.Then the value of $k$ is _______
Show Answer
Answer: 9
Solution:
Formula: Conditional Probability, Bayes’ Theorem
$E_1:$ Smokers
$P(E_1)=\frac{1}{4}$
$E_2:$ non-smokers
$P(E_2)=\frac{3}{4}$
$E$ : diagnosed with lung cancer
$P(E / E_1)=\frac{27}{28}$
$P(E / E_2)=\frac{1}{28}$
$P(E_1 / E)=\frac{P(E_1) P(E / E_1)}{P(E)}$
$=\frac{\frac{1}{4} \times \frac{27}{28}}{\frac{1}{4} \times \frac{27}{28}+\frac{3}{4} \times \frac{1}{28}}=\frac{27^{9}}{3 \sigma _{10}}=\frac{9}{10}$
$k=9$
Therefore, the value of $k$ is 9.
