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:

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

So using 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

Answer 2

0.03

Explanation:

The given question uses the concept of Conditional Probability. The general formula of conditional probability in terms of two events A and B is:

`P(A|B) = (P(A \cap B)/(P(B))`

In the given case, the two events are:

LC = Event that someone has lung cancer

S = Event that someone is smoker

The formula in terms of these events will be:

`P(LC | S)=(P(S \cap LC))/(P(S))`

Using the given values, we can find the probability that a random person is a smoker and has lung cancer P(S∩LC).

`0.158=(P(S \cap LC))/(0.19) \\\\ P(S \cap LC) = 0.158 * 0.19\\\\ P(S \cap LC) = 0.03`

Therefore, the probability (rounded to the nearest hundredth) that a random person is a smoker and has lung cancer P(S ∩ LC) is 0.03