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%)  

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

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step-by-step explanation:

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