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Question
A bag contains 3 yellow and 5 brown balls. Another bag contains 4 yellow and 6 brown balls. If one ball is drawn from each bag, what is the probability that, the balls are of different color?
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Solution
Let event A: A yellow ball is drawn from each bag.
Probability of drawing one yellow ball from total of 8 balls of the first bag and that of drawing one yellow ball out of total of 10 balls of the second bag is
P(A) = `(""^3"C"_1)/(""^8"C"_1)xx(""^4"C"_1)/(""^10"C"_1)`
= `3/8xx4/10`
= `3/20`
Let event B: A brown ball is drawn from each bag. Probability of drawing one brown ball out of total 8 balls of first bag and that of drawing one brown ball out of total 10 balls of second bag is
P(B) = `(""^5"C"_1)/(""^8"C"_1)xx(""^6"C"_1)/(""^10"C"_1)`
= `5/8xx6/10`
= `3/8`
Since both the events are mutually exclusive events, P(A ∩ B) = 0
∴ P(both the balls are of the same colour) = P(both are of yellow colour) or P(both are of brown colour)
= P(A) + P(B)
= `3/20+3/8`
= `3((2+5)/20)`
= `21/40`
P(both the balls are of different colour)
= 1 – P(both the balls are of the same colour)
= `1 - 21/40`
= `19/40`
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