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A bag contains some white and some black balls, all combination of balls being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are formed to be black. Let P(E) be the probability that bag contains 1 white and 9 black ball. Then $\frac{55}{2} P (E) =$

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answer is 7.

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Detailed Solution

No of black balls:	3	4	5	6	7	8	9	10
No of white balls:	7	6	5	4	3	2	1	0

(8 cases equally – likely)
Using Baye’s Theorems
P[box contains 9 black balls / 3 balls drawn are all black]
$= \frac{14}{55} = P (E)$
Now, $(\frac{55}{2}) P (E) = \frac{55}{2} \times \frac{14}{55} = 7$

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