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An urn contains 8 white and 4 red balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are white ? |
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Answer» Given that an urn contains 8 white and 4 red balls. Let experiment is drawn two balls from the urn one after the other without replacement. Therefore, total numbers of trials are n(S) = 12 × 11 = 132. Let the event is drawn two white balls from the urn. Therefore, the number of trials in which both drawn balls are white balls is n(A) = 8 × 7 = 56. Therefore, the probability that both drawn balls are white is \(\frac{n(A)}{n(s)}\) = \(\frac{56}{132}\) = \(\frac{14}{33}\) |
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