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Question
A bag contains 7 red and 4 black balls, 3 balls are drawn at random. Find the probability that one red and 2 black
Solution
Let S be the sample space
A be the event of taking 3 red balls and B be the event of taking one red and
2 black balls.
A bag contains 7 red balls and 3 black balls.
3 balls are drawn at random.
∴ The number of outcomes n(S) = 11C3
= `(11 xx 10 xx 9)/(1 xx 2 xx 3)`
= 11 × 5 × 3
= 165
n(A) = 7C3
= `(7 xx 6 xx 5)/(1 xx 2 xx 3)`
= 35
n(B) = 7C1 × 4C2
= `7 xx (4 xx 3)/(1 xx 2)`
= 7 × 6
= 42
P(getting one red and 2 blacks) =P(B)
= `("n"("B"))/("n"("S"))`
= `42/165`
= `14/55`
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