Three bags contain 3 red, 4 black balls; 4 red, 5 black balls; 5 red, 2 black balls respectively. If one bag is selected at random and a ball is drawn from it then the probability that it is red is 

Let A be the event of selecting the first bag, B be the event of selecting the second bag, C be the event of selecting the third bag and E be the event of drawing a red ball from the selected bag. 

$\therefore P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{3},P\left(C\right)=\frac{1}{3},P(E\mid A)=3/7,P(E\mid B)=4/9,P(E\mid C)=5/7$

$P\left(E\right)=P\left(A\right)P(E\mid A)+P\left(B\right)P(E\mid B)+P\left(C\right)P(E\mid C)$

$=\frac{1}{3}\times \frac{3}{7}+\frac{1}{3}\times \frac{4}{9}+\frac{1}{3}\times \frac{5}{7}=\frac{27+28+45}{189}=\frac{100}{189}$.