Probability: Red ball being transfefred from bag m to n?

[mirandab17]

Bag M contains 5 white balls and 2 red balls. Bag N contains 3 white balls and 4 red balls.
a) A ball is randomly selected from Bag M and Placed in Bag N. A ball is then
randomly selected from Bag N. What is the probability that the ball selected form
Bag N is white?
b) If a white ball is selected from Bag N, what is the probability that a red ball was
transferred from Bag M to Bag N?

a) I got this answer which is right: 13/28. I got this by taking the probablity that a white ball is chosen from bag m and then that a white ball is chosen from bag n (which now has one more white ball) and then multiplying that number by the probability of choosing a red ball from bag m, but then choosing a white ball which is the same, just one more red ball to choose from.

That sounds horribly complicated.
Overall: (5/7)(4/8) + (2/7)(3/8) = 13/28

b) This is where I get confused... I know it's a conditional probability problem, so GIVEN THAT the white ball is selected from bag n (my previous answer), aka (13/28) must be on the bottom... but how do I get what's on the top? A red ball transferred from Bag M to N... so I thought it must be (2/7)(5/8)...

PLEASEHELP!

 

[Ray Vickson]

Bag M contains 5 white balls and 2 red balls. Bag N contains 3 white balls and 4 red balls.
a) A ball is randomly selected from Bag M and Placed in Bag N. A ball is then
randomly selected from Bag N. What is the probability that the ball selected form
Bag N is white?
b) If a white ball is selected from Bag N, what is the probability that a red ball was
transferred from Bag M to Bag N?

a) I got this answer which is right: 13/28. I got this by taking the probablity that a white ball is chosen from bag m and then that a white ball is chosen from bag n (which now has one more white ball) and then multiplying that number by the probability of choosing a red ball from bag m, but then choosing a white ball which is the same, just one more red ball to choose from.

That sounds horribly complicated.
Overall: (5/7)(4/8) + (2/7)(3/8) = 13/28

b) This is where I get confused... I know it's a conditional probability problem, so GIVEN THAT the white ball is selected from bag n (my previous answer), aka (13/28) must be on the bottom... but how do I get what's on the top? A red ball transferred from Bag M to N... so I thought it must be (2/7)(5/8)...

PLEASEHELP!

Let W1 = first is white, R1 = first is red, W2 = second is white. You have computed P(W2) = P(W2|W1)P(W1) + P(W2|R1)P(R1). You want P(R1|W2). This is P(R1 & W2)/P(W2), and we can write P(R1 & W2) = P(W2 & R1) = P(W2|R1) P(R1). You know P(W2|R1), P(R1) and P(W2), so you have everything you need.

RGV

 

[mirandab17]

Let W1 = first is white, R1 = first is red, W2 = second is white. You have computed P(W2) = P(W2|W1)P(W1) + P(W2|R1)P(R1). You want P(R1|W2). This is P(R1 & W2)/P(W2), and we can write P(R1 & W2) = P(W2 & R1) = P(W2|R1) P(R1). You know P(W2|R1), P(R1) and P(W2), so you have everything you need.

RGV

Okay... I don't understand how P(R1&W2) turned into P(W2/R1) P (R1). Why are you multiplying it?

 

[Villyer]

A red ball transferred from Bag M to N... so I thought it must be (2/7)(5/8)...

Here you are saying that a red ball was picked from bag M and a red ball from bag N.
But the question states that a white ball was picked from bag N.

 

[mirandab17]

But it is conditional right? So IF a white ball was picked from bag N, which is 11/21, the probability of transferring a red ball from M to N is... I'm assuming red was picked first here then... so (2/7) and then white (3/8)...?

 

[mirandab17]

so (2/7)(3/8) / (11/21)?

 

[Villyer]

I think it's conditional. I don't know anything about probability outside of what I can deduce logically, but from what I've seen this matches that definition.

And your numerator is right now, but double check the fraction you placed in the denominator.

 

[mirandab17]

The denominator should be (3/7) because it's only involving bag N?

 

[Villyer]

b) This is where I get confused... I know it's a conditional probability problem, so GIVEN THAT the white ball is selected from bag n (my previous answer), aka (13/28) must be on the bottom...

You're original denominator was correct.