# Probability of picking buttons from a bag

A bag contains $$30$$ buttons that are colored either blue, red or yellow. There are the same number of each color ($$10$$ each). A total $$4$$ buttons are drawn from the bag. Compute the followings:

1. Find $$n(\Omega)$$.
2. The probability that at least $$3$$ of them are red?
3. The probability that there is at least one of each color?

This seems like a basic problem but my professor and I cannot agree on an answer.

I think the probabilities are $$2/21$$ and $$100/203$$ for parts $$2$$ and $$3$$ respectively. I used combinations to calculate the probabilities.

My professor said $$n(A)/n(\Omega)$$ is $$3/15$$ for both so that is the answer for both $$2$$ and $$3$$.

## 3 Answers

A bag contains $$30$$ buttons that are colored either blue, red or yellow. There is the same number of each color ($$10$$ each). A total of $$4$$ buttons are drawn from the bag. Compute the following: $$1$$. Find $$n(\Omega)$$.

$$2$$. The probability that at least $$3$$ of them are red.

$$\frac{{10\choose3}{20\choose1}+{10\choose4}}{{30\choose4}}=\frac{2610}{27405}=\frac{2\cdot3^2\cdot5\cdot29}{3^3\cdot5\cdot7\cdot29}=\frac 2{21}$$

$$3$$. The probability that there is at least one of each color.

$$\frac 12\times\frac{{10\choose1}{10\choose1}{10\choose1}{27\choose1}}{{30\choose4}}=\frac 12\times\frac{10\cdot10\cdot10\cdot27}{27405}=\frac 12\times\frac{2^3\cdot3^3\cdot5^3}{3^3\cdot5\cdot7\cdot29}=\frac {100}{203}$$ where the factor of $$\frac12$$ was to cancel double-counting, in that every combination like $$(\underbrace{R_i}_{{10\choose1}}\underbrace{B_i}_{{10\choose1}}\underbrace{Y_i}_{{10\choose1}}\underbrace{R_j}_{{27\choose1}})$$ is also counted in $$(R_jB_iY_iR_i)$$.

You are correct.

Part B can also be done as

$$P(B) = \frac{3C1.(10C2.10C1.10C1)}{30C4} = \frac{100}{203}$$.

• I did part B this way too! – Benjamin2002 Jul 25 '20 at 6:40
• @Benjamin2002 Actually, I knew that there is going to be a more straightforward way from experience. Instead of thinking about one, I thought, why not tell this approach that might help curb common mistakes like not taking that $\frac12$ factor. – Sameer Baheti Jul 25 '20 at 6:56

I think we have the same professor, and I just received my assignment back and I basically got all of it wrong because of this question, he said that it was 15 not 27405.

• Yeah, he said the sample space is 15 instead of 27405. I feel you man – Benjamin2002 Aug 1 '20 at 13:36