+207 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?
+1908 First, let's find the total number of ways Miyu and distribute the chocolates with no restrictions.
According to the Stars and Bars Theorem, we have
\(\binom{n + k - 1}{k - 1}\) where n is the identical items and k is the number of groups.
Plugging in 8 and 5, we have
\( \binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} = 495 \)
Next, we must find the number of favorable outcomes. If Druv gets 6 chocolates, then there are 2 more chocolates for 4 other friends.
Using the same method, we have
\( \binom{2 + 4 - 1}{4 - 1} = \binom{5}{3} = 10\)
Now, we can write the equation
\( P(\text{Dhruv gets 6}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{495}\)
So our answer is approximately 2.02%.
Thanks! :)
+1908 In that case, we can do the same with 7 and 8.
We have
\(\binom{1 + 4 - 1}{4 - 1} = \binom{4}{3} = 4\)
and
\(\binom{0 + 4- 1}{4 - 1} = \binom{3}{3} = 1\)
So there are 10+4+1 = 15.
So our answer is 15/495.
Thanks! :)
