let M and N be two events, if P(M) =0.12, P(N|M)=0.23, and P(N|M')=0.17

This can be done with formulas, but the thinking is easier with a Venn
diagram, using reduced sample spaces. [Formulas get you mixed up!]

Let the probabilities of the 4 regions be a,b,c,d.



P(M') = c+d
P(M|N) = b/(b+c)
P(M'|N) = c/(b+c)

P(M) = 0.12 = a+b
P(M') = c+d = 1-P(M) = 1-0.12 = 0.88
        c+d = 0.88

P(N|M) = 0.23 = b/(a+b) = b/0.12

        0.23 = b/0.12
(0.23)(0.12) = b 
      0.0276 = b

Substitute in
        0.12 = a+b
        0.12 = a+0.0276
      0.0924 = a

P(N|M') = 0.17 = c/(c+d) = c/0.88

        0.17 = c/0.88
(0.17)(0.88) = c
      0.1496 = c

Substitute in
        c+d = 0.88
   0.1496+d = 0.88  
          d = 0.7304

P(M|N) = b/(b+c) = 0.0276/(0.0276+0.1496) = 0.0276/0.1772 = 0.1557562077
P(M'|N) = c/(b+c) = 0.1496/(0.0276+0.1496) = 0.1496/0.1772 = 0.8442437923

Edwin