# Conditional probability and combinatorics question.

• peripatein
In summary: Again, the point is that the value of the highest seat number occupied by a woman depends on the order in which the participants are seated, and we are told the participants are seated randomly. So for example, if the women are seated in the first three seats, X is 1, but if they are seated in the last three seats, X is 4. You then calculated the probabilities for each possible value of X.
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?

## 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.

peripatein said:
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?

## 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.

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:
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?

peripatein said:
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.

## 1. What is conditional probability?

Conditional probability is a mathematical concept that calculates the likelihood of an event occurring given that another event has already occurred. It is represented by P(A|B), where P(A) is the probability of event A and P(B) is the probability of event B.

## 2. How is conditional probability different from regular probability?

Regular probability calculates the likelihood of an event occurring without taking into account any prior knowledge or information. Conditional probability, on the other hand, takes into consideration the occurrence of another event.

## 3. What is the formula for calculating conditional probability?

The formula for conditional probability is P(A|B) = P(A ∩ B) / P(B), where P(A ∩ B) is the probability of event A and B occurring together, and P(B) is the probability of event B occurring.

## 4. What is combinatorics in relation to conditional probability?

Combinatorics is a branch of mathematics that deals with counting and arranging objects. In relation to conditional probability, combinatorics is used to calculate the total number of possible outcomes in a given situation, which is then used to determine the probability of a specific event occurring.

## 5. Can you give an example of a conditional probability and combinatorics question?

Sure! Say you have a bag with 10 red marbles and 5 blue marbles. What is the probability of randomly selecting a red marble, given that a blue marble was previously selected and not replaced? This question involves both conditional probability (the probability of selecting a red marble given that a blue marble was previously selected) and combinatorics (calculating the total number of possible outcomes). The answer would be 10/14, or approximately 0.71.

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