# Simple probability problem

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1. Feb 28, 2016

### Titan97

1. The problem statement, all variables and given/known data
Find the probability of choosing one white ball from a bag containing 3 white balls and 2 red balls.
Find the probability of choosing 2 white balls from a bag containing 3 white balls and 2 red balls

2. Relevant equations
None

3. The attempt at a solution
For the first question, the answer is $\frac{3}{5}$

For the second question, since balls of one colour are identical, the number of ways of choosing any two balls is 3 because the possibilities are (ww),(wr),(rr). Or is it $\binom{5}{2}$? I am confused. Because in the first question, the total ways of choosing 1 ball is only 2 and not 5 since the outcomes are either (w) or (r). Then is my answer to first question wrong?!

2. Feb 28, 2016

### Ray Vickson

No, the number of ways of choosing any two balls is not three. Number the balls from 1 to 5. How many ways are there to choose two balls from those five?

3. Feb 28, 2016

### Titan97

Numbering the balls means you are considering them different right? But what if the balls were identical?

4. Feb 28, 2016

### Ray Vickson

No, they are not considered different, but the numbers are ignored when you look at the colors. The balls really are physically different objects.

5. Feb 28, 2016

### Titan97

In that case, the number of ways of choosing 2 balls is 5C2. But the number of ways of choosing R things from N identical objects is 1. Because no matter how you choose, the outcome is always same.

6. Feb 28, 2016

### Ray Vickson

It all depends on how you choose to do the counting, and why. The correct method could give different answers in different problems. In some problems you might be able to safely ignore the differences, and use your answer of 1. In other problems, the differences are crucial to correct counting, even if you ultimately ignore them in looking at the outcomes, and in that case the correct number of ways would be
$_NC_R$.

In your problem where there are 3 white and 2 red balls in an urn, and you draw two balls, the number of favorable outcomes (2 white) is $_3C_2 = 3$, while the total possible number of outcomes of all types is $N = {}_5C_2 = 10$, so the answer is P(2 white) = 3/10. If you don't believe this, we can do it another way that leaves no doubt. The probability that the first drawn ball is white is P1 = 3/5, because we are equally likely to draw any of the 3 whites from the total of 5 balls. Now, given that the first draw is white, we are left with 4 balls, of which 2 are white, so the probability of getting white on the second draw (given white on the first) is P2 = 2/4. The answer we want is the product of these two probabilities, which is (3/5)(2/4) = 6/20 = 3/10. The reason we multiply the probabilities is that we want P(W1 & WE), where W1 and W2 are the events W1 ={first is white} and W2 = {second is white}. A basic probability law is P(W1 & W2) = P(W1) * P(W2|W1), where P(W2|W1) is the conditional probability of W2, given that W1 has occurred.

Comparing these two ways of computing the answer, you can plainly see that using "1" for the number of ways of choosing 2 whites would give incorrect results, while using $_3C_2 = 3$ does it correctly. Again, the point is that we can, in principle, paint unique labels 1, 2 and 3 on the white balls, and then our favorable outcomes would be 12, 13 or 23.

Last edited: Feb 29, 2016
7. Feb 29, 2016

### haruspex

No, the number of ways of choosing R from N is NCR, but the number of effectively different results may be 1.

The key question is what are the atomic events, i.e. the finest grain of individual events that we can reasonably assume to be equally likely.
As Ray points out, two balls may be considered identical for the purposes of the result, but in the real world they are distinct objects, so choosing one rather than the other are two different atomic events.
If you choose one ball blindly from two white balls and one black, you do not equally likely get a white or a black simply because the outcome is the same. The equally likely events are choosing first ball, second ball, third ball.

A real world case of objects being truly indistinguishable is bosons in the same state, e.g. electrons. They obey different statistics.

8. Feb 29, 2016

### Ray Vickson

One final clarification: while it is true that the outcome is the same, there are several ways to get to that outcome; the probability of the outcome looks at the number of different ways of "getting there", not at the final destination itself.

9. Mar 2, 2016

### HallsofIvy

Staff Emeritus
Yes, good.

With or without replacement? If this is "with replacement" (take a ball out, note its color, then put it back and choose again) then the second time there are still 5 balls, 3 being white so the probability the second is also white is still 3/5 and the probability both balls are white is (3/5)(3/5)= 9/25.
If this is "without replacement" then after picking the first white ball, there are four ball left, two of the white so the probability the second ball is white is 2/4= 1/2 and the probability both balls are white is (3/5)(1/2)= 3/10.

Your reasoning here applies only if the two outcomes are "equally likely" and that is not the case here. Since there were different numbers of colored balls, "equally likely" applies to individual balls not colors.

10. Mar 3, 2016

### TheMathNoob

For this problem, I think it's better if you use conditional probability. If you draw the balls one at a time, then the logic of the events looks something like this
B= probability of getting a white ball in the first draw
A= Probability of getting a white ball in the second draw

So the expression above means that the probability of getting both a white ball in the first and second draw is equal to the probability of getting a white ball in the second draw given that you picked a white ball in the first draw times the probability of getting a white ball in the first draw

P(AlB)= 2/4
P(B)=3/5
P(AnB)= 1/2*3/5=0.3

11. Mar 3, 2016

See post #6.