• Support PF! Buy your school textbooks, materials and every day products Here!

Conditional probability and combinatorics question.

  • Thread starter peripatein
  • Start date
  • #1
880
0
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 [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.
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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 [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.
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.
 
  • #3
880
0
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?
 
Last edited:
  • #4
880
0
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?
 
  • #5
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,565
5,467
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.
 

Related Threads on Conditional probability and combinatorics question.

  • Last Post
Replies
2
Views
581
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
597
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
0
Views
853
Replies
5
Views
8K
Replies
1
Views
2K
Top