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

Answer 1

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`