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A bag contains 12 red balls 6 white bal...

A bag contains 12 red balls 6 white balls. Six balls are drawn one by one without replacement of which at least 4 balls are white. Find the probability that in the next two drawn exactly one white ball is drawn. (Leave the answer in `""^(n)C_(r )`).

Text Solution

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The correct Answer is:
A, B, C, D
Let `A_(1)` be the event exactly 4 white balls have been drawn. `A_(2)` be the event exactly 5 white balls have been drawn.
`A_(3)` be the event exactly 6 white balls have been drawn.
B be the event exactly 1 white ball is drawn from two draws. Then,
`P(B)=P((B)/(A_(1)))P(A_(1))+P((B)/(A_(2)))P(A_(2))+P((B)/(A_(3)))P(A_(3))`
But `P((B)/(A_(3)))=0`
[since, there are only 6 white balls in the bag]
`therefore P(B)=P((B)/(A_(1)))P(A_(1))+P((B)/(A_(2)))P(A_(2))`
`=(""^(12)C_(2)""^(6)C_(4))/(""^(18)C_(6))*(""^(10)C_(1)*""^(2)C_(1))/(""^(12)C_(2))+(""^(12)C_(1)*""^(6)C_(5))/(""^(18)C_(6))*(""^(11)C_(1)*""^(1)C_(1))/(""^(12C_(2)))`
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