Question

According to the American Lung Association, 7% of the population has lung disease. Of those having lung disease, 90% are smokers; and of those not having lung disease, 25% are smokers. What is the probability that a smoker has lung disease?

Answer 1

Answer:

Probability that a smoker has lung disease = 0.2132

Step-by-step explanation:

Let L = event that % of population having lung disease, P(L) = 0.07

So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93

S = event that person is smoker

% of population that are smokers given they are having lung disease, P(S/L) = 0.90

% of population that are smokers given they are not having lung disease, P(S/L') = 0.25

We know that, conditional probability formula is given by;

                        P(S/L) = $\frac{P(S\bigcap L)}{P(L)}$  

                        $P(S\bigcap L)$ = P(S/L) * P(L)

                                      = 0.90 * 0.07 = 0.063

So,  $P(S\bigcap L)$ = 0.063 .

Now, probability that a smoker has lung disease is given by = P(L/S)

      P(L/S) = $\frac{P(S\bigcap L)}{P(S)}$

P(S) = P(S/L) * P(L) + P(S/L') * P(L')

       = 0.90 * 0.07 + 0.25 * 0.93 = 0.2955

Therefore, P(L/S) = $\frac{0.063}{0.2955}$ = 0.2132

Hence, probability that a smoker has lung disease is 0.2132 .