Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**derek Deschler****Guest**

Four students are taking their “Applied Logic” final. Their names are Amy, Ben, Chad, and Dom. Their instructor tells them “I have four white hats and three black hats. I am going to blindfold you and place a hat on each of your heads. Then I’m going to line you up so you can only see the hats of the people in front of you. You won’t be able to see your own hat’s color. One at a time, starting from the back, I’ll ask you if you know what color your hat is. You can only respond by saying my hat is white, my hat is black, or I don’t know.” The instructor has the students blindfold themselves and he orders them with Amy in front, then Ben, then Chad, and finally Dom. He places hats on their heads. Once the students have their hats on, the instructor has them remove their blindfold. He goes to Dom and asks if she knows the color of her hat. Even though she can see the three other students, she doesn’t know what color her hat is, so she says “I don’t know.” The instructor the approaches Chad. Even knowing what Dom said and being able to see Ben and Amy’s hats, Chad also doesn’t know the color of his hat, so he says “I don’t know.” The instructor goes to Ben, the second person in line. Despite hearing that the students behind did not know the color of their hats and being able to see the hat on Amy’s head, Ben doesn’t know either and says “I don’t know.” Finally, Amy, even though she cannot see any of the other students hats and has only heard “I don’t know” from the other students, confidently says she knows the color of her hat. She is right. What color is Amy’s hat? How could she be certain she was correct?

i think i know the answer is white im just not sure how to explain it well enough to get a useful grade on it

**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 129

Hi Derek, it is straightforward for the final person to deduce the correct color of her hat. This can be proven as follows: For clarity let me call "the first person" from back (Dom) the first person and the next one "the second person" and so no. Since there are just three black hats so if the second, third and forth persons wearing them then the first person would be certain that his hat is white (remember that there are just three black hats and they know this fact). Therefore some of the hats on the other persons are black and some are while or all of them are white. The second person know the previous conditions from the answer of the first person, but if the two hats of the third and forth persons are black then he would be sure (from the information he had got from the first person and from his observation) that his hat is white, However, his uncertainty insured the third person and the forth person that not all of their hats are back. At this stage the third person would be sure that his hat most be white if the front person has black hat (since he know all the answers of the previous students) and the froth and final student "Amy" knew all of this information therefore she is certain that her hat is white.

Offline

**monie27****Member**- Registered: 2019-03-13
- Posts: 3

Any one good in business statistics?I am struggling with probability.

I have a one question that I am struggling with : the question is as follows.

The probability that a flight departs on time in 0.83 and the probability that it arrives on time is 0.92. The probability that the flight arrived on time given that it departed on time is 0.94 (rounded off to two decimals).

1.What is the probability that the flight did not arrive on time

2.What is the probability that the flight arrived and departed on time

3.What is the probability that the flight departed on time given that it has arrived on time?

4.What is the probability that the flight either arrived or departed on time

Offline

**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 129

Hi monie27, the answer is as follows: Let's the probabilities of arriving and departing on time are B and A, respectively. Then for the conditional probability, it can be proven that we have:

but we have

however since we have:

finally we have:

Therefore the answers are:

1-0.08

2-0.78

3-0.85

4-0.91

I hope that will help.

*Last edited by Grantingriver (2019-03-13 07:33:48)*

Offline

**monie27****Member**- Registered: 2019-03-13
- Posts: 3

Thank you very much Grantingriver ?

Offline