Probability difficult bowl of balls

[ArcanaNoir]

Homework Statement

A bowl contains w white balls. One ball is selected at random from the bowl; its color is noted, and it is returned to the bowl along with n additional balls of the same color. Another single ball is randomly selected from the bowl (now containing w+b-n balls) and it is observed that the ball is black. Show that the (conditional) probability that the first ball is selected was white is $\frac{w}{w+b-n}$

(this is exactly how it is written by hand by my prof)

Homework Equations

I'm using "n" in between A and B as "intersect"

Well, probability of a, given b, is P(A|B)= [P(AnB)]/[P(B)]
I think (for independent events) P(AnB)=P(A)xP(B)

The Attempt at a Solution

I guess to me it should be w plus b PLUS n, not minus. And I'm not sure how to use conditional probability here, it looks more independent to me. Halp!

 

[Dickfore]

Surely you mean that there are w + b + n balls in the bowl the second time, since you added n of them.

 

[ArcanaNoir]

That's what I'm ?? about, but the problem is definitely w+b-n.

What if there were w+b-2n balls, then when you add n, you have w+b-n. ??

 

[Dickfore]

no, it's a typo from that book you are checking all the problems here.

 

[ArcanaNoir]

Ordinarily I'd agree but this is not from the book, it's hand written problem made by my prof. But he has made mistakes too, so typo is highly possible. I think ill email him.

 

[I like Serena]

Hmm, I think I see another typo.
You write that initially the bowl contains w white balls.
How could a black ball then be drawn?


Btw, the equation I think you will need is Bayes' theorem:

 

[ArcanaNoir]

Hmm, I think I see another typo.
You write that initially the bowl contains w white balls.
How could a black ball then be drawn?


Btw, the equation I think you will need is Bayes' theorem:

Good call on the Bayes theorem, it's one of the key concepts of the homework. As for the balls, I think he's trying to get you to figure out what was in the bowl retroactively.

 

[Dickfore]

Hmm, I think I see another typo.
You write that initially the bowl contains w white balls.
How could a black ball then be drawn?


Btw, the equation I think you will need is Bayes' theorem:

Cause it had b black balls.

 

[I like Serena]

I give up!

This is not a probability problem, it's a find-the-typo problem, or a think-outside-of-the-box problem (since apparently we have to make weird assumptions that the bowl initially contains w white balls accompanied by an unknown number of other balls).

 

[ArcanaNoir]

I know, right? Well I'll update if prof gives a reasonable answer to my email.

 

[Ray Vickson]

Homework Statement

A bowl contains w white balls. One ball is selected at random from the bowl; its color is noted, and it is returned to the bowl along with n additional balls of the same color. Another single ball is randomly selected from the bowl (now containing w+b-n balls) and it is observed that the ball is black. Show that the (conditional) probability that the first ball is selected was white is $\frac{w}{w+b-n}$

(this is exactly how it is written by hand by my prof)

Homework Equations

I'm using "n" in between A and B as "intersect"

Well, probability of a, given b, is P(A|B)= [P(AnB)]/[P(B)]
I think (for independent events) P(AnB)=P(A)xP(B)

The Attempt at a Solution

I guess to me it should be w plus b PLUS n, not minus. And I'm not sure how to use conditional probability here, it looks more independent to me. Halp!

If W1 = {first ball is white} and B1 = {first ball is black}, what are P{W1} and P{B1}? Now, if {W2} and {B2} refer to the second ball, what are P{B2|W1}, P{B2|B1}, etc? You want P{W1|B2} = P{W1 & B2}/P{B2}. Can you get P{W1 & B2} in terms of P{W1}, P{B1}, P{B2|W1} and P{B2|B1}? Do you know how to get P{B2}?

By the way: after the first draw the number of balls *is* b + w + n, if your description of the experiment is correct.

RGV

 

[ArcanaNoir]

Okay, the prof just emailed me back. It is correct as w+b-n. It's due tomorrow. Can I get some more help on it?

 

[ArcanaNoir]

ILS, why do you say Baye's theorem is $$P(A|B)= \frac{P(B|A)P(A)}{P(B)}$$ ?

My book says $$P(A|B)= \frac{P(B|A)P(A)}{P(A)P(B|A)+P(B)P(B|A)}$$

 

[Dickfore]

ILS, why do you say Baye's theorem is $$P(A|B)= \frac{P(B|A)P(A)}{P(B)}$$ ?

My book says $$P(A|B)= \frac{P(B|A)P(A)}{P(A)P(B|A)+P(B)P(B|A)}$$

Because the denominator of both fractions is the same.

 

[ArcanaNoir]

How's that?

 

[Dickfore]

look at the formula for total probability in your book. It should be in the same section as Bayes's Theorem.

 

[ArcanaNoir]

I tried Baye's theorem and I got w/(w+b)
This seems kind of reasonable. Since the adding of the n balls occurs AFTER the drawing of the white ball. Now I see why there is -n. But I still have to get that for an answer.

 

[vela]

Did you ask your professor to clarify how it could be w+b-n? And for that matter, what does b stand for? Are we supposed to assume it's the number of black balls?

 

[ArcanaNoir]

Arg. After all that, during class today he admits that he wrote the problem inaccurately! Forget it.

 

[Dickfore]

lol.