Question

A recent survey by the cancer society has shown that the probability that someone is a smoker is P(S)=0.19. They have also determined that the probability that someone has lung cancer, given that they are a smoker is P(LC|S)=0.158. What is the probability (rounded to the nearest hundredth) that a random person is a smoker and has lung cancer P(S∩LC) ?

0.35


0.03


0.02


0.83

Answer 1

P (S∩LC) = 0.03

Explanation:

We are given that the probability that someone is a smoker is P(S)=0.19 and the probability that someone has lung cancer, given that they are a smoker is P(LC|S)=0.158.

Given the above information, we are to find the probability hat a random person is a smoker and has lung cancer P(S∩LC).

P (LC|S) = P (S∩LC) / P (S)

Substituting the given values to get:

0.158 = P(S∩LC) / 0.19

P (S∩LC) = 0.158 × 0.19 = 0.03