Conditional probability and combinatorics question.

[peripatein]

Hi,
I wish to confirm the results I obtained for the two following questions in statistics. I'd truly appreciate your feedback.

Homework Statement

1) Die 1 and die 2 form a pair of unbiased dice. Die 1 has 4 faces painted red and 2 painted blue, whereas die 2 has 4 faces painted blue and 2 painted red. A coin is tossed: in case 'heads' occurs, die 1 is rolled twice, and in case 'tails', die 2 is rolled twice. The coin is biased, so that 'heads' occurs with a probability of 3/4.
The event 'first rolling is blue' is termed A, and 'second rolling is blue' is termed B.
Are A and B independent of each other? Further, first rolling turned out to be red and the second turned out to be blue. What is the probability that die 2 was rolled?

2) 3 men and 3 women sit randomly in 6 chairs numbered 1-6. Let X be the lowest chair number in which a woman is seated. What would be the probability function X?

Homework Equations

The Attempt at a Solution

1) P(A) = 3/4 * 2/6 + 1/4 * 4/6 = 5/12
P(B) = 3/4 * 2/6 + 1/4 * 4/6 = 5/12
P(A AND B) = (3/4 * 2/6 + 1/4 * 4/6)(3/4 * 2/6 + 1/4 * 4/6) = 25/144 = P(A)*P(B), hence, A and B are independent.
P(die 2 | (rolling 1 red, rolling 2 blue)) = P(2 AND (rolling 1 red, rolling 2 blue )) / P(rolling 1 red, rolling 2 blue) = (1/18) / [(1/18) + (3/4*4/6*3/6)] = 1/4.

2) X can take any value between 1 and 4. I believe the sample space \Omega should be equal to 6!.
For X = 1, probability should be 3 * 5! / 6! = 1/2
For X = 2, probability should be [3 * (4! / 2!) * 3!] / 6! = 3/10
For X = 3, probability should be [3 * (3 * 2!) * 3!] / 6! = 3/20
For X = 4, probability should be [3 * (2!) * 3!] / 6! = 1/20

As mentioned, I'd sincerely appreciate your feedback on these attempts.

 

[Ray Vickson]

Hi,
I wish to confirm the results I obtained for the two following questions in statistics. I'd truly appreciate your feedback.

Homework Statement

1) Die 1 and die 2 form a pair of unbiased dice. Die 1 has 4 faces painted red and 2 painted blue, whereas die 2 has 4 faces painted blue and 2 painted red. A coin is tossed: in case 'heads' occurs, die 1 is rolled twice, and in case 'tails', die 2 is rolled twice. The coin is biased, so that 'heads' occurs with a probability of 3/4.
The event 'first rolling is blue' is termed A, and 'second rolling is blue' is termed B.
Are A and B independent of each other? Further, first rolling turned out to be red and the second turned out to be blue. What is the probability that die 2 was rolled?

2) 3 men and 3 women sit randomly in 6 chairs numbered 1-6. Let X be the lowest chair number in which a woman is seated. What would be the probability function X?

Homework Equations

The Attempt at a Solution

1) P(A) = 3/4 * 2/6 + 1/4 * 4/6 = 5/12
P(B) = 3/4 * 2/6 + 1/4 * 4/6 = 5/12
P(A AND B) = (3/4 * 2/6 + 1/4 * 4/6)(3/4 * 2/6 + 1/4 * 4/6) = 25/144 = P(A)*P(B), hence, A and B are independent.
P(die 2 | (rolling 1 red, rolling 2 blue)) = P(2 AND (rolling 1 red, rolling 2 blue )) / P(rolling 1 red, rolling 2 blue) = (1/18) / [(1/18) + (3/4*4/6*3/6)] = 1/4.

2) X can take any value between 1 and 4. I believe the sample space \Omega should be equal to 6!.
For X = 1, probability should be 3 * 5! / 6! = 1/2
For X = 2, probability should be [3 * (4! / 2!) * 3!] / 6! = 3/10
For X = 3, probability should be [3 * (3 * 2!) * 3!] / 6! = 3/20
For X = 4, probability should be [3 * (2!) * 3!] / 6! = 1/20

As mentioned, I'd sincerely appreciate your feedback on these attempts.

In 1) it makes a difference whether or not we know the results of the coin toss. What I mean is the following. In scenario (1) we toss the coin and we pick up the die to be tossed. In scenario (2) somebody else tosses the coin and hands us the die to be tossed, but without telling us what the coin-toss results are. The point is: in one of the two scenarios the events A and B are independent, but in the other scenario they are not.

 

[peripatein]

What about the rest of my answers?
And, mainly, how may I demonstrate, rigorously, that A and B are indeed dependent?
Moreover, is my answer (namely 1/4) to the second part of that question correct?

 

[peripatein]

I have meanwhile tried solving the first question, with the dice, again, and came up with the following results:
P(A) = P(B) = 5/12, whereas P(A AND B) = (2/6)*(2/6)*(3/4) + (4/6)*(4/6)*(1/4) = 7/36
Since 7/36 != 25/144(=5/12 * 5/12), A and B are dependent!
Ray, or anyone else willing to assist for that matter, would you kindly corroborate these numerical results?

 

[haruspex]

I have meanwhile tried solving the first question, with the dice, again, and came up with the following results:
P(A) = P(B) = 5/12, whereas P(A AND B) = (2/6)*(2/6)*(3/4) + (4/6)*(4/6)*(1/4) = 7/36
Since 7/36 != 25/144(=5/12 * 5/12), A and B are dependent!
Ray, or anyone else willing to assist for that matter, would you kindly corroborate these numerical results?

That's a correct analysis and result.
For the next part, yes it's 1/4. Each die has the same odds of producing one red and one blue, so knowing that outcomes tells you nothing about which die was rolled. Therefore it reduces to the chances of choosing die 2 in the first place.
You answered Q2 correctly.