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Question
Bag A contains 3 red and 2 white balls and bag B contains 2 red and 5 white balls. A bag is selected at random, a ball is drawn and put into the other bag, and then a ball is drawn from that bag. Find the probability that both the balls drawn are of same color
Solution
Let event C1: First ball drawn is red and from bag A,
event D1: First ball drawn is white and from bag A,
event E1: First ball drawn is red and from bag B,
event F1: First ball drawn is white and from bag B,
event C2: Second ball drawn is red and from bag B,
event D2: Second ball drawn is white and from bag B,
event E2: Second ball drawn is red and from bag A,
event F2: Second ball drawn is white and from bag A,
event G: Selecting bag A in first place,
event H: Selecting bag B in first place.
P(G) = P(H) = `1/2`
Let event X: Both the balls drawn are of same color.
∴ P(X) = `"P"("G") xx "P"("X"//"G") + "P"("H") xx "P"("X"//"H")` …(i)
If bag A is selected in first place, then In bag A, we have 5 balls, out of which 3 are red.
∴ Probability of getting first red ball from bag A = P(C1)
= `(""^3"C"_1)/(""^5"C"_1) = 3/5`
Since first red ball is put into the bag B, we now have 8 balls in bag B, out of which 3 are red.
∴ Probability of getting second red ball from bag B.
`"P"("C"_2//"C"_1) = (""^3"C"_1)/(""^8"C"_1) = 3/8`
Similarly, probability of getting first white ball from bag A = P(D1) = `(""^2"C"_1)/(""^5"C"_1) = 2/5` and probability of getting second white ball form bag B = `"P"("D"_2//"D"_1) = (""^6"C"_1)/(""^8"C"_1) = 6/8`
∴ `"P"("X"//"G") = "P"("C"_1) * "P"("C"_2//"C"_1) + "P"("D"_1) * "P"("D"_2//"D"_1)`
= `3/5 xx 3/8 + 2/5 xx 6/8`
= `21/40` ...(ii)
Similarly, `"P"("X"//"H")`
= `"P"("E"_1) * "P"("E"_2//"E"_1) + "P"("F"_1) * "P"("F"_2//"F"_1)`
= `2/7 xx 4/6 + 5/7 xx 3/6`
= `23/42` ...(iii)
From (i), (ii), (iii),
Required probability = `1/2 xx 21/40 + 1/2 xx 23/42`
= `3604/6720`
= `901/1680`
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