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

Answer:

P (S∩LC) = 0.03

Step-by-step 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

Answer:

0.03

Step-by-step 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:

$P(A|B) = \frac{P(A \cap B}{P(B)}$

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:

$P(LC | S)=\frac{P(S \cap LC)}{P(S)}$

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

$0.158=\frac{P(S \cap LC)}{0.19} \\\\ P(S \cap LC) = 0.158 \times 0.19\\\\ P(S \cap LC) = 0.03$

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