Miyu 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 at least $6$ chocolates?

Answer 1

To solve the problem, we will first determine the total number of ways to distribute the 8 identical chocolates among the 5 friends, and then we will find the number of ways in which Dhruv gets at least 6 chocolates.

Step 1: Total number of distributions

Let \( x_1, x_2, x_3, x_4, x_5 \) represent the number of chocolates received by each of the 5 friends (with \( x_1 \) corresponding to Dhruv). We need to find the number of non-negative integer solutions to the equation:

\[ x_1 + x_2 + x_3 + x_4 + x_5 = 8. \]

Using the stars and bars combinatorial method, the number of solutions is given by:

\[ \binom{n+k-1}{k-1} = \binom{8+5-1}{5-1} = \binom{12}{4}. \]

Calculating \( \binom{12}{4} \):

\[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11880}{24} = 495. \]

Step 2: Distributions where Dhruv gets at least 6 chocolates

Now we consider the scenario where Dhruv receives at least 6 chocolates. For this, we will examine two cases based on how many chocolates Dhruv gets:

1. Case 1: Dhruv gets exactly 6 chocolates. In this case, the remaining chocolates to be distributed to the other friends is \( 8 - 6 = 2 \). The problem becomes finding non-negative integer solutions to:

   \[ x_2 + x_3 + x_4 + x_5 = 2. \]

   Using stars and bars again, the number of solutions is:

   \[ \binom{2 + 4 - 1}{4 - 1} = \binom{5}{3} = 10. \]

2. Case 2: Dhruv gets exactly 7 chocolates. The remaining chocolates to be distributed is \( 8 - 7 = 1 \), and we need to solve:

   \[ x_2 + x_3 + x_4 + x_5 = 1. \]

   The number of solutions here is:

   \[ \binom{1 + 4 - 1}{4 - 1} = \binom{4}{3} = 4. \]

3. Case 3: Dhruv gets exactly 8 chocolates. In this case, there are no chocolates left for the other friends. The only solution is:

   \[ x_2 + x_3 + x_4 + x_5 = 0, \] which provides exactly 1 solution (i.e., \( x_2 = x_3 = x_4 = x_5 = 0 \)).

Total distributions where Dhruv gets at least 6 chocolates

Now we sum the solutions from all cases:

\[ 10 \text{ (for 6 chocolates)} + 4 \text{ (for 7 chocolates)} + 1 \text{ (for 8 chocolates)} = 15. \]

Step 3: Calculate the probability

The probability that Dhruv gets at least 6 chocolates is the number of favorable outcomes over the total number of outcomes:

\[ P(\text{Dhruv gets at least 6 chocolates}) = \frac{15}{495}. \]

We simplify this fraction:

\[ \frac{15}{495} = \frac{1}{33}. \]

Thus, the final answer is:

\[ \boxed{\frac{1}{33}}. \]