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A bag contains 3 yellow and 5 brown balls. Another bag contains 4 yellow and 6 brown balls. If one ball is drawn from each bag, what is the probability that, (a) both the balls are of the same colour? (b) the balls are of a different colours? |
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Answer» (a) Let event A: A yellow ball is drawn from each bag. Probability of drawing one yellow ball from total 8 balls of first bag and that of drawing one yellow ball out of total 10 balls of second bag is P(A) = \(\frac {^{3}C_1}{^{8}C_1}\)x \(\frac {^{4}C_1}{^{10}C_1} = \frac {3} {8} \) x \(\frac {4} {10} = \frac {3}{20}\) Let event B: A brown ball is drawn from each bag. Probability of drawing one brown ball out of total 8 balls of first bag and that of drawing one brown ball out of total 10 balls of second bag is P(A) = \(\frac {^{5}C_1}{^{8}C_1}\)x \(\frac {^{6}C_1}{^{10}C_1} = \frac {5} {8} \) x \(\frac {6} {10} = \frac {3}{8}\) Since both the events are mutually exclusive events, P(A ∩ B) = 0 ∴ P(both the balls are of the same colour) = P(both are of yellow colour) or P(both are of brown colour) = P(A) + P(B) = 3/20 + 3/8 = 21/40 (b) P(both the balls are of different colour) = 1 – P(both the balls are of the same colour) = 1- 21/40 = 19/40 |
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