Using conditional probability, it is found that there is a 0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.

Conditional Probability

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this problem:

Event A: Person has the flu.

Event B: Person got the flu shot.

The percentages associated with getting the flu are:

20% of 30%(got the shot).

65% of 70%(did not get the shot).

Hence:

$P(A) = 0.2(0.3) + 0.65(0.7) = 0.515$

The probability of both having the flu and getting the shot is:

$P(A \cap B) = 0.2(0.3) = 0.06$

Hence, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.515} = 0.1165$

0.1165 = 11.65% probability that a person with the flu is a person who received a flu shot.

To learn more about conditional probability, you can take a look at brainly.com/question/14398287

7 0

2 years ago