A bag of M&Ms and the probability of getting a red one

A bag contains 50 M&Ms; 10 of them are red.

(a) If half of the M&Ms are eaten, what is the probability that 5 red ones were among them?

My work:

P(5 red) = (((10 choose 5)(40 choose 20))/(50 choose 25) = 0.27

(b) If half of the M&Ms are eaten, Find the probability that 5 red ones are among the remaining M&Ms

My work:

P(red remaining) = ((10 choose 5)(15 choose 5))/(25 choose 10) = 0.23

(c) If half of the M&Ms are eaten, how many red M&Ms can you expect to be among them?

This part I had no idea how to do.


I am not sure if any of these are right or even close to being right. Can someone help explain all three parts?
Thank you!

Hello, bry32321!

A bag contains 50 M&Ms; 10 of them are red.

(a) If half of the M&Ms are eaten, what is the probability that 5 red ones were among them?

My work: \(\displaystyle P(\text{5 red eaten}) \:=\: \dfrac{{10\choose5}{40\choose20}}{{50\choose25}} \:=\:0.27\)

Click to expand...

I haven't checked your arithmetic, but your set-up is correct.

(b) If half of the M&Ms are eaten, find the probability that 5 red ones are among the remaining M&Ms.

My work: \(\displaystyle P(\text{5 red left}) \:=\: \dfrac{{10\choose5}{15\choose5}}{{25\choose10}} \:=\:0.23\)

Click to expand...

This is incorrect.

The answer should be the same as part (a).

(c) If half of the M&Ms are eaten, how many red M&Ms can you expect to be among them?

Click to expand...

This is an enormous problem!


. . \(\displaystyle \begin{array}{ccc} \text{Reds} && \text{Prob.} \\ \hline \\
0 & \times & \dfrac{{10\choose0}{40\choose25}}{{50\choose25}} \\ \\
1 & \times & \dfrac{{10\choose1}{40\choose24}}{{50\choose25}} \\ \\
2 & \times & \dfrac{{10\choose2}{40\choose23}}{{50\choose25}} \\ \\
\vdots && \vdots \\ \\ 10 & \times & \dfrac{{10\choose10}{40\choose15}}{{50\choose25}} \\ \\ \hline
\end{array}\)


Perform the multiplications and add the results.

soroban said:

This is an enormous problem!
Perform the multiplications and add the results.

Click to expand...

It does not have to be enormous. Look at this tool.

pka said:

It does not have to be enormous. Look at this tool.

Click to expand...

Even simpler - though perhaps feeling like waving hands -
If half of the population is randomly selected, the expectation for each color is half of its number.

E[red] = (1/2)(total red) = 5

Thank you very much everyone! Very helpful!