A leather box contains 8 black balls and 6 white balls. Two draws of three balls each are made, the balls being replaced after the first draw. What is the probability that the balls were black in the first draw and white in the second draw?

a) 70/8281 b)140/20449 c ) 25/5445 d)35/5448

• Total number of balls = 8 + 6 = 14
  Total ways of drawing 3 balls, N(S) = 14C3 = 364
  No of ways to draw 3 black balls = N(E1) for black balls = 8C3 = 56
  Probability of all balls being black = P(E1) = N(E1) / N(S) = 56/ 364 = 14/91
  No of ways to draw 3 white balls = N(E2) for white balls = 6C3 = 20
  Probability of all balls being white = P(E2) = 20 / 364 = 5/91
  Probability of drawing 3 black balls in first draw and drawing 3 white balls in the second draw = P(E1) x P(E2)
  Therefore P(E) = 14/91 x 5/91 = 70 /8281
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