Arecent 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

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:

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:

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

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

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

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p