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A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is(a) \(\frac{14}{99}\)(b) \(\frac{7}{99}\)(c) \(\frac{7}{35}\)(d) \(\frac{4}{12}=\frac{1}{3}\) |
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Answer» (b) \(\frac{7}{99}\) There are (7 + 5) = 12 balls in the bag. 4 balls can be drawn at random from 12 balls in 12C4 ways. ∴ n(S) = 12C4 = \(\frac{|\underline{7}}{|\underline3|\underline4}\) = \(\frac{7\times6\times5}{3\times2}\) = 35 ∴ Required probability = \(\frac{n(A)}{n(S)}\) = \(\frac{35}{495}\) = \(\frac{7}{99}\). |
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