# Balls in a box - probability

1. Dec 3, 2009

### duki

1. The problem statement, all variables and given/known data

A box has 10 bass, 6 are black and 4 are white. Three balls are removed from the box, color unknown. Find the probability that a fourth ball removed will be white.

2. Relevant equations

3. The attempt at a solution

I got:
$$\frac{4}{10} * \frac{3}{9} * \frac{2}{8} * \frac{1}{7} = 0.00476$$

2. Dec 3, 2009

How?

3. Dec 4, 2009

updated

4. Dec 4, 2009

### diazona

Ah, I see. You used 4/10, 3/9, and 2/8 as the probability for the first three balls - but those are the probabilities of picking all white balls. What if the first ball picked is black? Or if the second ball picked is black? Or if all of the first three are black? You haven't accounted for any of those possibilities.

You should first figure out how many different possibilities there are for the colors of the first 3 balls.

Then, once you've done that, for each of those possibilities, calculate the probability that the first three balls are those colors and then the fourth ball is white.

Finally, since those are mutually exclusive possibilities (the first three balls are either WWW, WWB, WBW, ... but only one of those), you can add up the probabilities for all the cases.

5. Dec 4, 2009

### duki

Ok, I got 0.3905?

6. Dec 4, 2009

### sylas

As phrased this is a trick question, I think; a very good one.

As phrased, it appears that you are given no information at all about the colour of the balls that are removed, right?

Think about this. Suppose you take two balls out of the box, in the dark, and put them in two paper bags. Now switch on the light, and look at your bags.

What is the probability that the ball in the first bag is white?
What is the probability that the ball in the second bag is white?

Cheers -- sylas

7. Dec 4, 2009

### diazona

That's not quite what I got. What did you add up? (Remember you have to show your work for as long as you continue to want help!)

I think sylas has in mind a different way of reasoning, trickier but shorter... I didn't mention that at first because I figured there was more educational value in doing it the straightforward way.

8. Dec 8, 2009

### zgozvrm

Without showing my work, I got 0.2.
Not sure if I'm right though.

9. Dec 8, 2009

### sylas

I got twice as much as that....

10. Dec 8, 2009

### zgozvrm

After re-calculating, so did I!