Q. 84.8( 6 Votes )

# An urn contains 3

There are 3 white, 4 red and 5 black balls in the bag, so the number of all favorable outcomes in the sample space is

n(S) = 3+4+5=12

Let A be the event of not getting a black ball in the first draw, this means getting another color (red or white) ball out of 12 balls in the first draw. Hence the probability becomes

(as there are 3 white + 4 red = 7 balls)

Let B represents the event of not getting a black ball in the second draw, this means getting another color (red or white) ball out of 11 balls in the second draw as the balls are not replaced. Hence the probability becomes

(as there are 3 white + 4 red = 7 balls and one ball is already drawn in first draw so now there are total of 6 red and white balls)

Then the probability of at least one black ball without replacement

Hence the required probability is

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