# Two similar Bayesian problems. Did I get them right?

1. Mar 12, 2006

### Fredrik

Staff Emeritus
I would appreciate if someone could help me solve and understand these two problems. The first version of this post contained my attempts to solve them, but I have deleted those parts because I saw that I had messed up badly. I really suck at this type of problems, but perhaps someone can show me a good way to approach them.

Edit: OK, now I think I get it. If I still think I'm right in 20 minutes I'll probably edit this post again and add my new attempts to solve the problems.

Another edit: I have to go to bed, so I don't have time to post the explanations, but the results I get are 10/11 for the first problem and 1/2 for the second. Does that sound right?

This is not homework by the way. Oh, and since I'm asking about "Bayesian" probabilities, I would also appreciate if someone could tell me how that word is supposed to be pronounced. Baysian? Buy-eeshan? Buy-eezian? Bay-eezian? I've been wondering about that for years.

Problem 1 Two identical boxes. One of them contains 10 balls numbered 1-10. The other one contains 100 balls numbered 1-100. You don't know if the box on the left contains 10 or 100. You use a coin flip to choose one of the boxes and ask a friend to pick a ball at random from it. The ball he picks has the number 9 written on it. What is the probability that the box you chose contained 10 balls?

Problem 2 Two identical buildings. Both of them contain 100 rooms numbered 1-100. 110 people are blindfolded and randomly put into rooms 1-10 of building A, and rooms 1-100 of building B. You're one of those people, and you're told that your room number is 9. What's the probability that you're in building A?

Last edited: Mar 12, 2006
2. Mar 12, 2006

### AKG

"Bayesian" comes from a guy named "Bayes", which would be pronounced like "Bays". So "Bayesian" is pronounced like "Bays-Ian". You can actually here it here. Anyways, the idea is to use the idea of conditional probability:

$$P(A | B) = \frac{P(A \cap B)}{P(B)}$$

So the probability of A, given that B has occured, is equal to the probability that both A and B occur, divided by the probability that B occurs. In problem 1, you want to calculate the probability that the box chosen had 10 balls, given that the ball that was picked had 9 written on it.

3. Mar 13, 2006

### Fredrik

Staff Emeritus
Thanks AKG. At least now I know how pronounce Bayesian. I'm not sure I understand the formula for the conditional probability though. A and B are obviously not independent in the formula. If they were, the formula would be kind of pointless, since we would have P(A|B)=P(A) and P(A and B)=P(A)*P(B). What I don't understand is what P(A and B) means when A and B are not independent.

Anyway, that doesn't matter much right now, since I believe my solutions are correct. If I'm wrong, I hope someone will tell me.

These are my solutions:

Problem 1

There was a 1/2 probability that you picked the box with 10 balls and a 1/2 probability that you picked the box with 100 balls. If you picked the box with 10 balls, it was certain that the ball your friend picked would have a number less than 11. If you picked the box with 100 balls, there was only a 1/10 chance that he would pick a ball with a number less than 11, and a 9/10 chance that he would not. From this we get the probabilities for each possibility:

. . . . . . . . . . . . . Small number . . . . . . . . . Large number

Box with 10 . . . . . 1/2 * 1 = 1/2 . . . . . . . . 1/2 * 0 = 0

Box with 100 . . . . 1/2 * 1/10 = 1/20 . . . . . . 1/2 * 9/10 = 9/20

If we do this a large number of times, we will get a ball with a small number from the box with 10 balls 1/2 the times and we will get a small number 1/2 + 1/20 = 11/20 times. The probability we seek is the first of those numbers divided by the second: (1/2)/(11/20)=10/11.

Problem 2

You were assigned a room that was chosen at random from a set of 110 rooms, 10 of which is in building A, so there was a 1/11 probability that you ended up in building A and a 10/11 probability that you ended up in building B. If you ended up in building A, it was certain that you would get a low room number. If you ended up in building B, there was a 1/10 probability that you would get a low room number and a 9/10 probability that you would get a high room number. From this we get the probabilities for each possibility:

. . . . . . . . . . . . . Small number . . . . . . . . . Large number

Building A . . . . . . 1/11 * 1 = 1/11 . . . . . . . 1/11 * 0 = 0

Building B . . . . . . 10/11 * 1/10 = 1/11 . . . . 10/11 * 9/10 = 9/11

If you do this a large number of times, you will find yourself in a room with a small number in building A 1/11 times, and you will find yourself in a room with a small number 2/11 times. The probability we seek is the first of those numbers divided by the second: (1/11)/(2/11)=1/2.

4. Mar 13, 2006

### DavidK

I would suggest that you use the formula:

$$P(A|B)= \frac{P(B)P(B|A)}{P(A)}$$

instead. In the first example $$A$$ is "the box contains 10 balls" and $$B$$ is "You pick ball number 9", we get $$P(B)= \frac{1}{2}(\frac{1}{10}+\frac{1}{100})$$, $$P(B|A)= \frac{1}{10}$$, and $$P(A)=\frac{1}{2}$$. Hence $$P(A|B)= \frac{10}{11}$$.

5. Mar 16, 2006

### EnumaElish

Did you mean P(A|B) = P(A)P(B|A)/P(B)?

6. Mar 17, 2006

Yes I did

7. Nov 9, 2008