# Probability: batch of 3 chips

1. Feb 11, 2016

### Les talons

1. The problem statement, all variables and given/known data
A bowl contains 3 red, 4 green, and 5 blue chips. A batch of 3 chips is withdrawn at random.
What is the probability the batch contains only one color?
What is the probability the batch contains exactly two colors?

2. Relevant equations

3. The attempt at a solution
P(all red) = 3/12 *2/11 *1/10
P(all green) = 4/12 *3/11 *2/10
P(all blue) = 5/12 *4/11 *3/10
What does it mean if the batch has only one color? Should I put the three probabilities above, or do I have to multiply them together?
P(2 red) = 3/12 *2/11 *9/10
P(2 green) = 4/12 *3/11 *8/10
P(2 blue) = 5/12 *4/11 *7/10
Again, what does it mean if a batch has exactly two colors? Should I put three possible answers?

2. Feb 11, 2016

### Samy_A

You multiply probabilities when you have independent events and you want to compute the probability that all these independent events occur. This is not the case here.

What is greater: P(all red) or P(batch contains only one color)?
Another question: what is the relation between (all red)∪(all green)∪(all blue) and (batch contains only one color)
I'm not sure why you compute these probabilities.
I would look at P(batch contains all three colors) to start with.

3. Feb 11, 2016

### Les talons

That's what I don't follow; since there are three cases for batch has one color.
P(all red) < P(all green) < P(all blue)
Is the probability the batch is all one color the P(all red) because it is the smallest of these cases?

(all red)U(all green)U(all blue) is a disjoint union of 3 sets? So the probability the batch has one color is the sum of the individual probabilites(?) (batch contains one color) = (all red)U(all green)U(all blue)

Why look at P(all 3 colors in batch)?
red 3, green 4, blue 5
P(red,green,blue) = 3/12 *4/11 *5/10
P(red,blue,green) = 3/12 *5/11 *4/10
= P(green,red,blue)
= P(green,blue,red)
= P(blue,red,green)
= P(blue,green,red)

4. Feb 11, 2016

### Samy_A

No.
Yes! Your explanation is also correct.
There are three possibilities:
E1: only 1 color in batch.
E2: precisely 2 colors in batch.
E3: all 3 colors in batch.

So P(E1)+P(E2)+P(E3)=1
P(E1) has been computed in the first part of the exercise.
P(E2) is what you have to compute.
If you can compute P(E3), then you can get P(E2) from P(E2)=1-P(E1)-P(E3)

(That's how I would do it. I'm sure there are other ways to get this.)
So you get 6*(3*4*5/(12*11*10)) for P(E3).

5. Feb 11, 2016

### HallsofIvy

Staff Emeritus
What you are doing here is correct. There are, initially, 12 chips, 3 of which are red so the probability of the first chosen being red is 3/12= 1/4. After that happens there are 11 chips left, 2 of which are red so the probability of the second chosen being red is 2/11. After that happens there are 10 chips left one of which is red. The probability of the third chosen being red is 1/10. The probability of three things A, B, and C happening is P(A)P(B/A)P(C/A and B)- the probability of A times the probability of A given B times the probability of C given that A and B have happened. That is, you multiply them together as you have: P(all red)= (3/12)(2/11)(1/10). Similarly for "all green" and "all blue".

"has only one color" means "all red or all blue or all green". The probability of "A or B or C", as long as A, B, and C are mutually exclusive is P(A)+ P(B)+ P(C). Since if we pick "all red" we cannot pick "all green" or "all blue" these are "mutually exclusive so you add these numbers.

The probability the first chip chosen is 3/12= 1/4 as before. If that happens there are 11 chips left, two of which are red so, as before, the probability the second chosen is red is 2/11. If that happens there are 10 chips left, 9 of which are NOT red so the probability the third chosen is not red is 9/10.
HOWEVER, (3/12)(2/11)(9/10) is the probability of "red, red, not red" in THAT order. There are two other orders: "red, not red, red" and "not red, red, red". The probability of "two red, one not red" in any order is 3(3/12)(2/11)(9/10).

Similarly, each of these must be multiplied by 3.

Since these cases are "mutually exclusive", you add the three.

6. Feb 11, 2016

### Les talons

Very well explained, thanks guys, I understand this better, but another question, "When we account for order above, because there are 3 ways to select for example, red, red, red'; red, red', red; red', red, red; this means to consider all the possible choices in combinations?" So it is equivalent to multiplying by 3. This is because the 3 choices are each three possible outcomes?

7. Feb 11, 2016

### Ray Vickson

A quick way of doing this type of problem is to recognize it a involving the "multiclass hypergeometric" distribution; specifically, a 3-class situation in your case.

Given a batch of $N$ items of three types ($N_1$ of type 1, $N_2$ of type 2 and $N_3$ of type 3), we choose a sample of $n$ items without replacement. Then, the probability of having $k_1$ items of type 1, $k_2$ items of type 2 and $k_3$ items of type 3 in the sample is
$$P(k_1,k_2,k_3) = \displaystyle \frac{{N_1 \choose k_1} {N_2 \choose k_2} {N_3 \choose k_3}}{{N \choose n}}$$
for any $k_1 + k_2 + k_3 = n$,
You have $N_1 = 3$ (red), $N_2 = 4$ (green), $N_3 = 5$ (blue). Thus,
$$P(k_1,k_2,k_3) = \frac{{3 \choose k_1} {4 \choose k_2} { 5 \choose k_3} }{{12 \choose 3} }= \frac{1}{220} {3 \choose k_1} {4 \choose k_2} {5 \choose k_3}$$

For the first question you want $P_1 =P(3,0,0) + P(0,3,0) + P(0,0,3)$. For the second question you want $P_2 = [P(1,2,0) + P(2,1,0)] + [P(1,0,2) + P(2,0 1)] + [P(0,1,2) + P(0,2,1)]$, but as pointed out in #4, an easier way would be to get $P_3 = P(1,1,1)$ and then get $P_2 = 1-P_1 -P_3$.

Last edited: Feb 11, 2016