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A box contains 6 white and 4 black marbles. Three marbles are drawn at random from this box. Find the probability that they are: (a) two white and one black marbles. (b) one white and two are black marbles. |
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Answer» please send answer today Box contains 6 white and 4 black marbles. \(\therefore\) Number of ways selecting 3 marbles from 10 marbles is n(s) = 10C3 = \(\frac{10.9.8.7!}{3.2.7!}\) = 120 (a) Number of ways of selecting 2 white and 1 black marble from 6 white and 4 black marbles = 6C2 x 4C1 \(\therefore\) n(E1) = \(\frac{6.5}2\times4=60\) \(\therefore\) Probability that selecting 3 marbles are 2 white and 1 black marble is P(E1) = \(\frac{n(E_1)}{n(S)}=\frac{60}{120}\) = 1/2 (b) Number of ways of selecting 1 white and 2 black marble is P(E2) = \(\frac{n(E_2)}{n(S)}=\frac{36}{120}\) = 6/20 = 3/10 |
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