A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now,  $P\frac{A}{E}$ = Probability of drawing a red ball when the first bag has been selected = 4/7

   $P\frac{A}{E}$  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) =  $PE1\times P\frac{A}{E}+PE2\times P\frac{A}{E}$ 

                          =  $\frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{2}{6}=\frac{19}{42}$