Question

In a certain population of women, 4 percent have had breast cancer, 20% are smokers, and 3% are smokers and have had breast cancer. What is the probability that a woman selected at random has had cancer or smokes, or both? Report with accuracy to 2 decimals (as a proportion, not percentage).

Answer 1

Answer:

$P(B)= 0.04 , P(S) = 0.2 , P(B \cap S) =0.03$

We want to calculate this probability:

$P(B \cup S)$

And we can use the total probability rule and we have this:

$P(B \cup S) = P(B) +P(S) -P(B \cap S) = 0.04+0.2- 0.03= 0.21$

The probability that a woman selected at random has had cancer or smokes, or both is 0.21

Step-by-step explanation:

For this case we define the following events

B= the women selected had breast cancer

S= the women selected is a smoker

$B \cap S$= the women selected is a smoker and had breat cancer

And for this case we have the following probabilities given:

$P(B)= 0.04 , P(S) = 0.2 , P(B \cap S) =0.03$

And we want to calculate this probability:

$P(B \cup S)$

And we can use the total probability rule and we have this:

$P(B \cup S) = P(B) +P(S) -P(B \cap S) = 0.04+0.2- 0.03= 0.21$

The probability that a woman selected at random has had cancer or smokes, or both is 0.21