12 marbles are selected at random from a large collection of

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Here is what the question is essentially asking: Imagine four bins—white, red, green, and yellow. We are distributing 12 identical transparent marbles to these bins. A marble, when it gets to the bin, takes on that bin's color. The question asks: if these 12 marbles are distributed randomly into the bins, what is the probability that at least one marble gets into each bin?

When distributing identical objects—12 marbles—we should use the "Stars and Bars" method.

"Stars and Bars" method:

Imagine the 12 marbles as 12 stars in a row. To divide these marbles among 4 bins, we need 3 bars to create 4 separate sections. For example:

***|***|***|*** would represent that the first bin got 3 marbles , the second got 3, the third got 3, and the fourth got 3.||**********|** would represent that the first bin got 0 marbles, the second got 0, the third got 10, and the fourth got 2.****||*****|*** would represent that the first bin got 4 marbles, the second got 0, the third got 5, and the fourth got 3.

So, for the total number of cases, the problem becomes one of arranging these 12 identical stars and 3 identical bars in a row, which is given by 15!/(12!3!) = 455.

For the case when at least one marble gets into each bin, we are essentially distributing only the remaining 12 - 4 = 8 marbles. Similarly, the problem becomes one of arranging these 8 identical stars and 3 identical bars in a row, which is given by 11!/(8!3!) = 165.

Therefore, the probability is 165/455 = 33/91.

Answer: D.

P.S. Check other questions related to distribution in DISTRIBUTING ITEMS/PEOPLE/NUMBERS... (QUESTION COLLECTION) from our Special Questions' Directory.­
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