Miyu is giving out 9 identical chocolates to her 7 friends, including Dhruv. All possible distributions are equally likely. What is the probability that Dhruv gets exactly 2 chocolates?
We can use the stars and bars method to count the number of ways to distribute the 9 identical chocolates among Miyu's 7 friends. Imagine 9 stars (representing the chocolates) and 6 bars (representing the 7-1 = 6 separations between the friends). We can place the bars in any of the 9+6 = 15 positions, and the chocolates will be distributed accordingly. For example, if we place the first bar in the second position, then the first friend gets 1 chocolate, and so on.
The number of ways to distribute the chocolates is therefore:
(9+6) choose 6 = 15 choose 6 = 5005
Each of these distributions is equally likely. To count the number of distributions in which Dhruv gets exactly 2 chocolates, we can fix two of the chocolates for Dhruv, and distribute the remaining 7 chocolates among the other 6 friends. We can use the same method as before, with 7 stars and 5 bars. The number of distributions in which Dhruv gets exactly 2 chocolates is therefore:
(7+5) choose 5 = 12 choose 5 = 792
Therefore, the probability that Dhruv gets exactly 2 chocolates is:
792 / 5005 ≈ 0.158
So the probability is approximately 0.158, or about 15.8%.