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peripatein

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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 [itex]\Omega[/itex] 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.