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

[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.

 

[Dick]

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.

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.

 

[GreenPrint]

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

 

[Dick]

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

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.

 

[haruspex]

(B)

What is the actual question for B?

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

That would not be true in general. $P(B_{1} \wedge R) = P(R)P(B_{1}|R)$, $P(B_{1}) = P(R)P(B_{1}|R) + P(\neg R)P(B_{1}|(\neg R)$

 

[GreenPrint]

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?
¬?

 

[HallsofIvy]

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 #

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

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.

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

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 $P(B_1)= P(R)P(B_1|R)+ P(W)P(B_1|W)$)

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

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.

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

 

[haruspex]

What is that symbol mean?
¬?

It's a logical 'not'.