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Question
Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn from Jar I, and if it is Tail, a ball is drawn from Jar II. Suppose that this experiment is done and a white ball was drawn. What is the probability that this ball was in fact taken from Jar II?
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Solution
Let event J1: Ball drawn from jar I,
event J2: Ball drawn from jar II.
P(J1) = P(head) = `1/2`
P(J2) = P(tail) = `1/2`
Let event W: Ball drawn is white.
In Jar I, there are total 12 balls, out of which 5 balls are white.
∴ Probability that the ball drawn is white under the condtion that it is drawn from Jar I.
= `"P"("W"//"J"_1)`
= `(""^5"C"_1)/(""^12"C"_1)`
= `5/12`
Similarly, `"P"("W"//"J"_2)`
= `(""^3"C"_1)/(""^15"C"_1)`
= `3/15`
= `1/5`
Required probability = `"P"("J"_2//"W")`
By Bayes’ theorem
`"P"("J"_2//"W") = ("P"("J"_2) * "P"("W"//"J"_2))/("P"("J"_1) * "P"("W"//"J"_1) + "P"("J"_2) * "P"("W"//"J"_2))`
= `(1/2 xx 1/5)/(1/2 xx 5/12 + 1/2 xx 1/5)`
= `(1/5)/((25 + 12)/60`
= `12/37`.
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