12 marbles are selected at random from a large collection of

I'm not getting the partition part , what do you mean by this ?

Total cases:

\(x_1+x_2+x_3+x_4=12…C(n+r-1,r-1)=(12+4-1,4-1)=\frac{15!}{3!12!}\)

Favorable cases:

\(x_1+1+x_2+1+x_3+1+x_4+1=12…x_1+x_2+x_3+x_4=12-4=8\)

\(C(n+r-1,r-1)=(8+4-1,4-1)=\frac{11!}{3!8!}\)

Probability: Favorable/Total

\(\frac{\frac{11!}{3!8!}}{\frac{15!}{3!12!}}=\frac{11!(3!12!)}{3!8!(15!)}=\frac{11!12!}{8!15!}=\frac{33}{91}\)

Ans (D)

­Hi chetan2u , I believe you used \((n+r-1)C(r-1)\) but this can be used when n identical items are distributed to r individuals. Are the balls here identical? Please help me to understand. Thank you in advance.

In this case the r distinct individuals are the distinct colors.

The identical items are the numbers, meaning not that 2=3😀, but that 2 of one color is the same number, 2, of a different color.

It's a bit nuanced, yes.

­Okay, so basically after the distribution of 4 marbles, the remaining (12-4)=8 marbles are distributed among 4 different colours (which is represented by 'r' i.e. individuals). So the formula I mentioned applies. This is what my understanding is as per what you said. Please correct me if I am wrong. Thank you­

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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 distribuition in DISTRIBUTING ITEMS/PEOPLE/NUMBERS... (QUESTION COLLECTION) from our Special Questions' Directory.­­
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@jackfr2Given: 12 marbles are selected at random from a large collection of white, red, green and yellow Marbles. The number of Marbles of each colour is unlimited.
Asked: Find the probability that the selection contains at least one marble of each colour?
Let the unlimited number of marbles for each color is a very large number L.

Favorable ways in which 1 marbles of each color is already distributed.
W + R +  G + Y  = 8
Number of favorable ways = (8+4-1)C(4-1) = 11C3 = 165

Total ways 
W + R +  G + Y  = 12
Number of favorable ways = (12+4-1)C(4-1) = 15C3 = 455

The probability that the selection contains at least one marble of each colour = 165/455 = 33/91

IMO D
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