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Question
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
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Solution
Let the events of choosing the first and second bag be E1 and E2 respectively, then
P(E1) = P(E2) = `1/2`
Let E be the event of drawing red balls, then
P(E|E1) = `4/8 = 1/2`, P(E|E2) = `2/8 = 1/4`
Now, what is the probability of drawing a ball from the first bag when it is known that it is red in colour.
= P(E1|E)
By Bayes' theorem P(E1|E) = `(P(E_1) xx P(E|E_1))/(P(E_1) xx P(E|E_1) + P(E_2) xx P(E|E_2))`
= `(1/2 xx 1/2)/(1/2 xx 1/2 + 1/2 xx 1/4)`
= `(1/4)/(1/4 + 1/8)`
= `1/4 xx 8/3`
= `2/3`
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