1. ## Binomial Distribution/Probability

Hi all,

I'm having trouble with figuring out how to solve problems similar to the one below:

Suppose the proportion of marbles that we get is as follows: 2 red, 2 green, 1 yellow and 1 blue. What is the probability that we'll get 3 red marbles out of 5?

To get the probability, I tried to find the binomial probability:

n = 5, r = 3, p = 0.4, q = 0.6

5!/3!2! x 0.4^3 x 0.6^2 = 0.2304

I'm told though that .2304 is not the right answer but I'm not sure why this is.

Can anyone guide me to the right direction (is it really incorrect?)?

2. ## Re: Binomial Distribution/Probability

Originally Posted by tardis
Hi all,

I'm having trouble with figuring out how to solve problems similar to the one below:

Suppose the proportion of marbles that we get is as follows: 2 red, 2 green, 1 yellow and 1 blue. What is the probability that we'll get 3 red marbles out of 5?

To get the probability, I tried to find the binomial probability:

n = 5, r = 3, p = 0.4, q = 0.6

5!/3!2! x 0.4^3 x 0.6^2 = 0.2304

I'm told though that .2304 is not the right answer but I'm not sure why this is.

Can anyone guide me to the right direction (is it really incorrect?)?
Where did you get p = 0.4?

2/6 of the marbles are red, so...

3. ## Re: Binomial Distribution/Probability

Ignore this. Totally wrong

4. ## Re: Binomial Distribution/Probability

Ah... I think I've been staring at the problem for too long without realizing my mistake. Thank you!

5. ## Re: Binomial Distribution/Probability

Hello, tardis!

Very clumsy wording . . .

There is a bag with 2 red, 2 green, 1 yellow and 1 blue marble.
We draw 5 marbles with replacement.
What is the probability that we will get 3 red marbles?

The bag contains: 2 Reds and 4 Others.

So we have: . $\begin{Bmatrix}P(\text{Red}) &=& \frac{2}{6} &=& \frac{1}{3} \\ \\[-4mm] P(\text{Other}) &=& \frac{4}{6} &=& \frac{2}{3} \end{Bmatrix}$

$P(\text{3 Red, 2 Others}) \;=\;{5\choose3}\left(\frac{1}{3}\right)^3\left( \frac{2}{3}\right)^2 \;=\;10\cdot\frac{1}{27}\cdot\frac{4}{9} \;=\;\frac{40}{243}$

6. ## Re: Binomial Distribution/Probability

Agh! So sorry. I glanced too quickly at the problem and should have seen you were using 2 out of 5 instead of 2 out of 6 Even though its already correctly answered here is the correct table for completeness sake:

Total Tries: 7776
0: 1024 0.131687
1: 2560 0.329218
2: 2560 0.329218
4: 320 0.0411523
5: 32 0.00411523