

InterviewSolution
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Bag I contains 8 white and 7 black balls and bag II contains 5 white and 4 black balls. One ball is randomly transferred from bag I to bag II. Then a ball is drawn from bag II. Find the probability that the ball so drawn is white. |
Answer» Give that bag I contains 8 white and 7 black balls and Bag II contain 5 white and 4 black balls. One ball from I bag randomly put in II bag. So the possibility is that ball taken out from bag I is white. Then probability that white ball is choosen from bag I = 8/15 Now total number of balls in bag II = 5 + 1 = 6 Probability that white ball is choosen from bag II = 6/10 When both events occurs together ∴ Probability = 8/15 × 6/10 = 48/150 Another possibility is that ball is choosen from bag I is black. Then probability that black ball is choosen from bag = 7/15 Now number of black ball in bag II = 4 + 1 = 5 ∴ Probability that black ball is choosen = 5/10 Probability that both events happen together = 7/15 × 5/10 = 35/150 ∵ Both events are mutually exclusive so only one event can happen. ∴ Required probability = 48/150 + 35/120 = 83/150 |
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