Box I contains 5 red and 6 black marbles. Box II contains 4 red and 5 black marbles. One marble is drawn at random from Box I and dropped into Box II. A marble is then drawn at random from Box II and dropped into Box I. A marble is now picked at random from Box I. Find the probability that the last marble drawn from Box I is red.
:
C
A red marble can be drawn from Box I afterthe two rounds of transfers in the following four ways:
Case 1: R→R→R
Probability = (511×5110×511)
Case 2: R→B→R
Probability = (511×5110×411)
Case 3: B→B→R
Probability = (611×6110×511)
Case 4: B→R→R
Probability = (611×6110×611)
Thus resulting probability = sum of probabilities of individual cases
=(511×5110×511) +(511×5110×411) +(611×6110×511) +(611×6110×611) = 6211210
Therefore, answer option (c) is the correct answer choice.
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