Answer:
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%)
