# 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)**