# Probability Question - colored balls in 2 bowls - Baye's formula?

1. Feb 4, 2013

### GreenPrint

Probability Question -- colored balls in 2 bowls -- Baye's formula?

Bowl #1 contains 2 red balls and 2 white balls
Bowl #2 contains 3 red balls and 2 white ball

one bowl is chosen at random (each is equally likely)

(A) What is the probability of choosing a red ball?

Let R stand for choosing a red ball
$B_{#}$ = Bowel #

$P(R) = P(B_{1})P(R|B_{1})+P(B_{2})P(R|B_{2})$
$P(R) = \frac{1}{2} \frac{1}{2} + \frac{1}{2} + \frac{3}{4} = .625$

(B)
$P(B_{1}) = P(R)P(B_{1}|R)$

I don't know how to calculate $P(B_{1}|R)$. Do I need to use Baye's formula?

Thanks for any help! Hopefully I'm not making this to complicated again lol.

Last edited by a moderator: Feb 4, 2013
2. Feb 4, 2013

### Dick

Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

I'm not an expert in these things but you got part (A) wrong. Try it again. And for part (B), now it sounds like a good time to use Baye's formula.

Last edited: Feb 4, 2013
3. Feb 4, 2013

### GreenPrint

Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

Is it just (1/2)(3/4) for part A?

4. Feb 4, 2013

### Dick

Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

No, no. It's given by exactly the formula you wrote except you've got P(R|B_2) wrong. What is that? It's not 3/4.

5. Feb 4, 2013

### haruspex

Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

What is the actual question for B?
That would not be true in general. $P(B_{1} \wedge R) = P(R)P(B_{1}|R)$, $P(B_{1}) = P(R)P(B_{1}|R) + P(\neg R)P(B_{1}|(\neg R)$

6. Feb 4, 2013

### GreenPrint

Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

Ok so I figured out my mathematical miscalculation in A and was able to figure out the answer to B (I realized I forgot to post it).

What is that symbol mean?
¬?

7. Feb 5, 2013

### HallsofIvy

Staff Emeritus
Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

Surely you don't mean that second "+"?
$$\frac{1}{2}\frac{1}{4}+ \frac{1}{2}\frac{3}{5}= \frac{1}{8}+ \frac{3}{10}$$
Ignoring your typo with the "+", the probability that, drawing from the second bowl, you will draw a red ball, is 3/(3+ 2)= 3/5, not 3/4.

What, exactly, is the question? This is simply a "formula". What are you to do with it?
(And, by the way, an incorrect formula. After having picked a single ball, the probability that it was picked from bowl 1 is $P(B_1)= P(R)P(B_1|R)+ P(W)P(B_1|W)$)

That at least is easy to answer. There are a total of 5 red balls, 2 of them from bowl 1. Given that you picked a red ball, the probability it came from bowl 1 is 2/5.

8. Feb 5, 2013

### haruspex

Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

It's a logical 'not'.