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

Probability that a smoker has lung disease = 0.2132

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) =
`(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) =
`(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) =
`(0.063)/(0.2955)` = 0.2132

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