An urn contains 6 black balls, 4 white balls and 5 red balls. The balls are removed 1 at a time fromthe urn and not replaced. If 3 balls are removed, what is the probability that:(a) All three are red.(b) Two are black and one is white(c) All are the same colour(d) At least 1 is white
Question
An urn contains 6 black balls, 4 white balls and 5 red balls. The balls are removed 1 at a time fromthe urn and not replaced. If 3 balls are removed, what is the probability that:(a) All three are red.(b) Two are black and one is white(c) All are the same colour(d) At least 1 is white
Solution
(a) All three are red.
The total number of balls in the urn is 6 (black) + 4 (white) + 5 (red) = 15 balls.
The probability of drawing a red ball on the first draw is 5/15 = 1/3.
Since the balls are not replaced, the total number of balls decreases by 1 after the first draw. So, the probability of drawing a red ball on the second draw is 4/14 = 2/7.
Similarly, the probability of drawing a red ball on the third draw is 3/13.
Therefore, the probability that all three balls are red is (1/3) * (2/7) * (3/13) = 6/273 ≈ 0.022.
(b) Two are black and one is white
The probability of drawing a black ball on the first draw is 6/15.
The probability of drawing a black ball on the second draw is 5/14.
The probability of drawing a white ball on the third draw is 4/13.
Therefore, the probability that two balls are black and one is white is (6/15) * (5/14) * (4/13) = 120/2730 ≈ 0.044.
However, the white ball can be drawn at any of the three draws. So, we multiply this probability by 3 (for the three different orders the balls can be drawn in), giving us a final probability of 0.132.
(c) All are the same colour
This is the sum of the probabilities that all three balls are black, all three balls are white, and all three balls are red.
We've already calculated the probability that all three balls are red: 0.022.
The probability that all three balls are black is (6/15) * (5/14) * (4/13) = 120/2730 ≈ 0.044.
The probability that all three balls are white is (4/15) * (3/14) * (2/13) = 24/2730 ≈ 0.009.
Therefore, the probability that all three balls are the same colour is 0.022 + 0.044 + 0.009 = 0.075.
(d) At least 1 is white
The easiest way to calculate this is to find the probability that none of the balls are white, and then subtract this from 1.
The probability that the first ball is not white is 11/15.
The probability that the second ball is not white is 10/14.
The probability that the third ball is not white is 9/13.
Therefore, the probability that none of the balls are white is (11/15) * (10/14) * (9/13) = 990/2730 ≈ 0.363.
So, the probability that at least one ball is white is 1 - 0.363 = 0.637.
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