A bag contains 4 red and 3 black balls A second bag contains 2 red and 4 black balls One bag is sele...........

Explanation:

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(AE1)  = Probability of drawing a red ball when the first bag has been selected = 4/7

  P(AE2) = 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) = P(AE1) x P(A/E₁) + P(AE2) x P(A/E₂)

= 1/2 x 4/7 + 1/2 x 2/6 = 19/42

Hence, option 2 is the correct answer.