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
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
- P(B|A) is the probability of event B happening, given that A happened.
- [tex]P(A \cap B)[/tex] 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:
[tex]P(A) = 0.2(0.3) + 0.65(0.7) = 0.515[/tex]
The probability of both having the flu and getting the shot is:
[tex]P(A \cap B) = 0.2(0.3) = 0.06[/tex]
Hence, the conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.515} = 0.1165[/tex]
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 https://brainly.com/question/14398287