An urn contains 10 red and 12 blue balls. They are withdrawn one at a time...

An urn contains 10 red and 12 blue balls. They are withdrawn one at a time without replacement until a total of 4 red balls have been withdrawn. Find the probability that exactly 7 balls withdrawn/

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Expert Solution

The R shows the red ball and B shows the blue ball. Since we need to draw 7 balls such that last ball is red and out of first 6 balls 3 are also red.

Out of first six balls, 3 must be red and 3 blue so possible number of ways we can draw 3 red balls and 3 blue balls is

Let us calculate the probability of sequence RRRBBBR. Out of 10+12 = 22 balls, 10 are red so probability of getting first red balls is 10/22. After that 9 red balls are remaining out of remaining 21 balls. So the probability of getting second red balls is 9 / 21.Likewise probability of above sequence is

P(RRRBBBR) = (10/22) * (9/21) * (8/20) *(12/19) * (11/18) * (10/17) * (7/16) = (10*9*8*7*12*11*10) / (22*21*20*19*18*17*16)

The probability of each possible 20 sequence will be same as above. So the probability that exactly 7 balls withdrawn is

20 * (10*9*8*7*12*11*10) / (22*21*20*19*18*17*16) = 0.1548

Answer: 0.1548

milcah answered 7 months ago

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