Colored Marbles
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Tim, Ted and Tom each have a bag containing five different colored marbles. Each bag contains the same five colors.
Tim reaches into Ted's bag and Tom's bag and randomly withdraws a marble from each and places them in his bag. Ted then reaches into Tim's bag and Tom's bag and randomly removes a marble from each and places them in his bag. Finally, Tom reaches into Tim's bag and Ted's bag and randomly chooses a marble from each and places them in his bag.
What is the probability that each has five different colors of marbles in their bag?
HintIt's simpler to work the problem from both ends.
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Answer
13/315 ~ 0.04127
Tim either draws the same color marble from each bag (1/5 probability) or a different color from each bag (4/5 probability.) So after Tim draws two marbles, he either has three of one color and one of each of the other four colors, or two of each of two colors and one of each of the other three colors. Both Ted and Tom have four different colored marbles.
When Tom draws a marble from each bag at the end, he will have three different colored marbles. In order for each person to end up with five different colors of marbles, when Tom draws each of the other two need to have two marbles of one color and one marble of each of the other four colors. Any other arrangement of marbles can't result in each person having five marbles after Tom draws. Tom will need to draw one of the two marbles that is the same color from each bag. There is a 1/3 chance of Tom drawing the correct color marble from each bag and a (1/3) * (1/3) = 1/9 probability of drawing a correct marble from both bags.
If Tim has three marbles of one color, then Ted will need to pick one of those marbles in order to have at least one of each color. There is a 3/7 probability of drawing one of these three.
If Tim has two marbles of each of two colors, then Ted will need to draw one of those four marbles. If Ted draws a marble of the color he is missing (2/7 probability), then he can draw any marble from Tom. If Ted draws one of the marbles of the color he already has (2/7 probability), then he will need to draw the color he is missing from Tom (1/4 probability).
The probability is then
((1/5)*(3/7) + (4/5)*(2/7 + (2/7)*(1/4))) * (1/9)
= (3/35 + (4/5)*(2/7 + 1/14)) * (1/9)
= (3/35 + (4/5)*(5/14)) * (1/9)
= (3/35 + 2/7) * (1/9)
= (13/35) * (1/9)
= 13/315
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