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Peeyuu is giving out 8 identical chocolates to her 5 friends, including Dhruv. All possible distributions are equally likely. What is the probability that Dhruv gets exactly 1 chocolate?

Thanks a whole lot for the help guys...

Dec 19, 2023

#2
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Imagine Dhruv getting 1 chocolate and the remaining 7 chocolates being distributed among the other 4 friends.

We can represent this with 7 bars and 1 star. The star represents Dhruv's chocolate, and the bars separate the chocolates for the other friends.

There are a total of 12 positions for the star and bars (7 bars + 1 star = 8). However, some arrangements are indistinguishable.

For example, if Dhruv gets the first chocolate and his friends get the next 7 in a specific order, it's the same as if he gets the second chocolate and his friends get the next 7 in the same order.

To count the indistinguishable arrangements, we need to consider the number of ways to arrange the 7 bars.

There are 7! ways to arrange 7 distinct objects, but we overcount by a factor of 4! because the 4 friends are indistinguishable.

Therefore, there are 7!/4! = 210 indistinguishable arrangements.

The total number of possible distributions for the 8 chocolates is 5^8 (each friend can get any of the 5 chocolates).

Therefore, the probability that Dhruv gets exactly 1 chocolate is the number of favorable arrangements divided by the total number of arrangements: 210 / 5^8

Dec 24, 2023