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A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. |
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Answer» Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively. Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) = P (C) = P (D). Since out of 16 balls, 11 are not white, therefore, P (A) = \(\frac{11}{16}\). ∴ Required probability = P (A) . P (B) . P (C) . P (D) = \(\frac{11}{16}\) x \(\frac{11}{16}\) x \(\frac{11}{16}\) x \(\frac{11}{16}\) = \(\big(\frac{11}{16}\big)^4.\) |
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