Pick a white, then a black, then a white

[BMAD]

Three bags contain 6 marbles each; all white or black: one bag has 5 white and 1 black, another has 4 white and 2 black, the third has 3 white and 3 black. One

white marble is drawn from one bag and one black marble is drawn from another (it is not known which bags). What is the probability of drawing a white marble from the remaining bag?

[superprismatic]

.68 precisely.

[bonanova]

To solve, we need to know if the first two picks were done [a] blindly or Monte Hall entered the room, looked into a bag unknown to us and removed a white marble, then went to another bag, unknown to us, looked inside and grabbed a black marble and then handed us the third bag and asks us now to guess its contents (calculate p of drawing white).

I'm thinking we should assume it's [a] and go talk with *gulp* Professor Bayes.

In that case an acceptable paraphrase of OP is: There are three bags (as described) someone grabbed a bag and with eyes closed pulled out a marble. It turned out to be white. Then he grabbed one of the other bags and pulled out, eyes closed, another marble. It turned out to be black. Setting those bags aside, we are handed the third bag and asked to say the probability of drawing white.

[bonanova]

p(white) = 50/76 = 0.65789, a little less than 2/3.

Professor Bayes advises that ...

1. The probability we are picking from Bag 1 is just
   the probability the initial white and black picks were
   B2B3 or B3B2 = 4/12 3/4 + 3/12 2/3 = 5/12
   
2. Probability we're picking from Bag 2 is
   probability of W/B being taken from
   B1B3 or B3B1 = 5/12 3/5 + 3/12 1/3 = 4/12
   
3. Prob we're picking from Bag 3 is
   probability W/B were initially taken from
   B1B2 or B2B1 = 5/12 2/5 + 4/12 1/4 = 3/12

The probability of picking a white from bags B1 B2 B3 = 5/6, 4/6, 3/6 respectively.

Multiplying pairwise and adding, we get the overall probability

p(white) = 5/12 5/6 + 4/12 4/6 + 3/12 3/5 = 50/76.

[BMAD]

p(white) = 50/76 = 0.65789, a little less than 2/3.

Professor Bayes advises that ...

1. The probability we are picking from Bag 1 is just

   the probability the initial white and black picks were

   B2B3 or B3B2 = 4/12 3/4 + 3/12 2/3 = 5/12

2. Probability we're picking from Bag 2 is

   probability of W/B being taken from

   B1B3 or B3B1 = 5/12 3/5 + 3/12 1/3 = 4/12

3. Prob we're picking from Bag 3 is

   probability W/B were initially taken from

   B1B2 or B2B1 = 5/12 2/5 + 4/12 1/4 = 3/12

The probability of picking a white from bags B1 B2 B3 = 5/6, 4/6, 3/6 respectively.

Multiplying pairwise and adding, we get the overall probability

p(white) = 5/12 5/6 + 4/12 4/6 + 3/12 3/5 = 50/76.

The bags are picked at random but there is a flaw in your logic.

[BMAD]

.68 precisely.

Care to share your solution?

[superprismatic]

The probability of picking one of each color from bags 1&2 is 14/36.

The probability of picking one of each color from bags 1&3 is 18/36.
The probability of picking one of each color from bags 2&3 is 18/36.
So, by Bayes, the probability of having used bags 1&2 initially, given that one of each color was chosen is 14/50.
So, by Bayes, the probability of having used bags 1&3 initially, given that one of each color was chosen is 18/50.
So, by Bayes, the probability of having used bags 2&3 initially, given that one of each color was chosen is 18/50.
Therefore, the remaining bag is 3 with probability 14/50;
the remaining bag is 2 with probability 18/50;
the remaining bag is 1 with probability 18/50;
So, the probability of picking a white from the remaining bag is (14/50)*(1/2)+(18/50)*(2/3)+(18/50)*(5/6)
which equals .68 exactly.