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Question
A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls are made without replacement. Find the probability that the first drawing will give 3 white balls and second drawing will give 3 red balls.
Solution
Total number of balls = 8 + 5 = 13
3 balls can be drawn out of 13 balls in 13C3 ways.
∴ n(S) = 13C3
Let A be the event that all 3 balls drawn are white.
3 white balls can be drawn out of 5 white balls in 5C3 ways.
∴ n(A) = 5C3
∴ P(A) = `("n"("A"))/("n"("S")) = (""^5"C"_3)/(""^13"C"_3) = (5 xx 4 xx 3)/(13 xx 12 xx 11) = 5/143`
After drawing 3 white balls which are not replaced in the bag, there are 10 balls left in the bag out of which 8 are red balls.
Let B be the event that the second draw of 3 balls are red.
∴ Probability of drawing 3 red balls, given that 3 white balls have been already drawn, is given by
`"P"("B"/"A") = (""^8"C"_3)/(""^10"C"_3) = (8 xx 7 xx 6)/(10 xx 9 xx 8) = 7/15`
∴ Required probability = P(A ∩ B)
= `"P"("A")."P"("B"/"A")`
= `5/143 xx 7/15`
= `7/429`
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