# 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 - Mathematics and Statistics

Sum

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?

#### 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.

Concept: Baye'S Theorem
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