70% of kids who visit a doctor have a fever, and 40% of kids who visit a doctor have a sore throat. If the probability that a kid has a sore throat given that he has a fever is 30%, what is the probability that he has a fever given that he has a sore throat?

52.5%

Step-by-step explanation:

Let A be kids who have a fever, and B be kids who have a sore throat, so

Based on the given information we have

P (A) = 70%, P (B) = 40%, and P (B|A) = 30% (B|A is read as "B given A", which means probability of B happening given that A has already happened)

We are asked to find P (A|B), the probability of a kid having a fever given that we already know he has a sore throat.

Step 1: First find P (A and B), this is the probability of a kid having both a fever and a sore throat. We use the formula

P (B|A) = P (A and B) / P (A)

We have 2 of these values listed above, so we plug them in ...

30% = P (A and B) / 70%

This give us a value for P (A and B) which equals 21% [multiply both sides by 70% to isolate P (A and B), (30%) (70%) = 21%]

Now we flip the equation to P (A|B), which is

P (A|B) = P (A and B) / P (B)

We have 2 of the values, so we plug them in ...

P (A|B) = 21%/40%

This gives us a value of 52.5% (divide 21% by 40%)