Question

In an effort to find the source of an outbreak of food poisoning at a conference, a team of medical detectives carried out a study. They examined all 50 people who had food poisoning and a random sample of 200 people attending the conference who didn’t get food poisoning. The detectives found that 40% of the people with food poisoning went to a cocktail party on the second night of the conference, while only 10% of the people in the random sample attended the same party. Which of the following statements is appropriate for describing the 40% of people who went to the party? (Let F = got food poisoning and A = attended party.)

a. P(F|A) = 0.40

b. P(A|FC) = 0.40

c. P(F|AC) = 0.40

d. P(AC|F) = 0.40

e. P(A|F) = 0.40

Short answer

The required answer is:

$P(A\mid F)=0.40$

Step-by-step solution

Given information

$40\%$of the people with food poisoning are going to a cocktail party

Let,$F$= has Food Poisoning and$A$= Attended party

Calculation

According to the given statement, there is a $40\%$chance that a person has good poisoning and attended the party.

Or equivalently: Given $F,$there is a $40\%$probability that A occurs.

After the vertical line, the "given" event must be given, and the other event must be mentioned before the vertical line. As a result, the conditional probability can be expressed as:

$P(A\mid F)=40\%=0.40$

It is observed that the correct option is$\left(e\right)$