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avatar+148 

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?

 Jun 17, 2024
 #1
avatar+670 
+1

There are 187 ways.

 Jun 17, 2024
 #2
avatar+1075 
-1

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! :)

 Jun 17, 2024
 #3
avatar+8 
-1

My queen, it says at least 6 chocolates. 

 

YASSSSSSSSSS!

slayqueen  Jun 17, 2024
edited by slayqueen  Jun 17, 2024
 #4
avatar+1075 
+1

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! :)

NotThatSmart  Jun 19, 2024
edited by NotThatSmart  Jun 19, 2024

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