Hi,

Phy 14

Cal 29

     43 (3 both) That is: 11 + 26 + 3 = 40

only Calculus 26 

P(only calculus) = 26/40 = 13⁄20   

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
 
The other tutor is correct. However let's use the formulas:

P(A OR B) = P(A) + P(B) - P(A AND B)

P(X) = P(X and Y) + P(X and Y')

and a little more detail, in case you need it.

We will assume that all 40 are taking either physics or calculus or both,
and that there are no students of the 40 who are not taking at least one
of them.  (Otherwise there would be more than one solution).
So the probability that they are taking one or the other or both
is assumed to be 1.

There are four categories the 40 students can be in

#1. Those taking physics but not calculus.
#2. Those taking BOTH physics and calculus.
#3. Those taking calculus but not physics.  <-- what we are looking for!
#4. Those NOT taking either physics or calculus.

     P(calculus) = P(#2) + P(#3)

A)         29/40 = P(#2) + P(#3)

P(physics OR calculus) = P(physics) + P(calculus) - P(physics AND calculus)

               1 = 14/40 + 29/40 - P(#2)

Multiply through by 40

              40 = 14 + 29 - 40P(#2)
              40 = 43 - 40P(#2)
         40P(#2) = 3
           P(#2) = 3/40

Substitute in equation A)

A)         29/40 = P(#2) + P(#3)
           29/40 =  3/40 + P(#3)
           26/40 = P(#3)
           13/20 = P(#3) 

Edwin
 

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