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.
Solution
Correct Answer: Option B
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, \(P\left(\frac A{E_1}\right)\) = Probability of drawing a red ball when the first bag has been selected = 4/7
\(P\left(\frac A{E_2}\right)\) = 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) = \(P\left(E_1\right)\times P\left(\frac A{E_1}\right)+P\left(E_2\right)\times P\left(\frac A{E_2}\right)\)
= \(\frac12\times\frac47+\frac12\times\frac26=\frac{19}{42}\)
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