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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.
23/42
19/42
7/32
16/39
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.
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