A bag contains 3 white, 3 black & 2 red balls. If three balls are drawn without replacement what is the probability that the third ball is red?

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The bag contains 3 white, 3 black and 2 red balls. 3 balls are drawn without replacement. For the 3rd ball to be red, the first two cannot both be red.

If the first ball picked is red, the probability of this happening is 2/8, the second ball can be either black or white with a probability of 6/7, the third can be red with a probability of 1/6. Multiplying the individual probabilities gives (2*6*1)/(8*7*6). If the second ball is red the probability of the third being red is (6*2*1)/(8*7*6). If only the third ball is red the probability of this happening is (6*5*2)/(8*7*6)

The sum of (2*6*1)/(8*7*6), (6*2*1)/(8*7*6) and (6*5*2)/(8*7*6) is 1/4.

This gives the probability of the third ball being red as (1/4)

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