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 )`).

A bag contains 12 red balls 6 white balls. Six balls are drawn one by

1.	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 )`).
Answer» Correct Answer - `(""^(10)C_(1)xx""^(2)C_(1))/(""^(12)C_(2))xx(""^(12)C_(2)xx""^(6)C_(4))/(""^(18)C_(6))+(""^(11)C_(1)xx""^(1)C_(1))/(""^(12)C_(6))xx(""^(12)C_(1)xx""^(6)C_(5))/(""^(18)C_(6))` Let us define the following events `{:(A:4,"white balls are drawn in first six draws"),(B:5,"white balls are drwn in fiest six draws"),(C:6,"white balls are drawn in first six draws"),(E:, "exactly one white ball is drawn in next two draws"):}` `" "("i.e., one white and one red")` Then `P(E)=P(E//A)P(A)+P(E//B)P(B)+P(E//C)P(C)` But `P(E//C)=0" "["as there are only 6 white balls in the bas"]` `thereforeP(E)=P(E//A)P(A)+P(E//B)P(B)` `=(""^(10)C_(1)xx""^(2)C_(1))/(""^(12)C_(2))(""^(12)C_(2)xx""^(6)C_(4))/(""^(18)C_(6))+(""^(11)C_(1)xx""^(1)C_(1))/(""^(12)C_(1))(""^(12)C_(1)xx""^(6)C_(5))/(""^(18)C_(6))`

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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 )`).

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