+77 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...
+1768 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
