A conditional probability question on picking out chips from a bowl.

[Hodgey8806]

Homework Statement

You are a member of a class of 18 students. A bowl contains 18 chips: 1 blue and 17 red. Each student is to take 1 chip from the bowl without replacement. The student who draws the blue chip is guaranteed an A for the course.

a)If you have the choice of drawing first, fifth, or last, which position would you choose? Justify you choice on the basis of probability.

b) Suppose the bowl contains 2 blue and 16 red chips. Which position would you now choose?

Homework Equations

I use a combinatorics formula that I will show on solution.

The Attempt at a Solution

I said that the 1st person has a simple chance of winning: 1/18

The 5th person, however, has to wait for 4 others to select the wrong chip so:
[(1C0)(17C4)/(18C4)] * (1/14). This gives me the probability of 1/18 though.

And the last person has to wait for 17 people to pick the wrong chip:
[(1C0)(17C17)/(18C17)]*(1/1). This give me the probability of 1/18 as well.

So I said I would like to be the 5th person for another reason. Last person has an unlikely chance that the people before him would not pick the blue chip (1/18). The 1st person has a (1/18) shot in the dark. While there is a 7/9 chance that the other people will pick the wrong chip. Leaving him a 1/14 chance of picking the blue chip.

I will pick up on part b later if the first part is correct. Does what I did make sense?

 

[Office_Shredder]

So I said I would like to be the 5th person for another reason. Last person has an unlikely chance that the people before him would not pick the blue chip (1/18). The 1st person has a (1/18) shot in the dark. While there is a 7/9 chance that the other people will pick the wrong chip. Leaving him a 1/14 chance of picking the blue chip.

So you have to balance the fact that you have a lesser chance of being able to pick with the higher chance of picking the right chip when it's your turn. To decide whether that's a good thing or not, you calculate the probability of winning. And you found that all the positions had the same probability. So which spot is the best? They're all exactly the same!

 

[Ray Vickson]

Homework Statement

You are a member of a class of 18 students. A bowl contains 18 chips: 1 blue and 17 red. Each student is to take 1 chip from the bowl without replacement. The student who draws the blue chip is guaranteed an A for the course.

a)If you have the choice of drawing first, fifth, or last, which position would you choose? Justify you choice on the basis of probability.

b) Suppose the bowl contains 2 blue and 16 red chips. Which position would you now choose?

Homework Equations

I use a combinatorics formula that I will show on solution.

The Attempt at a Solution

I said that the 1st person has a simple chance of winning: 1/18

The 5th person, however, has to wait for 4 others to select the wrong chip so:
[(1C0)(17C4)/(18C4)] * (1/14). This gives me the probability of 1/18 though.

And the last person has to wait for 17 people to pick the wrong chip:
[(1C0)(17C17)/(18C17)]*(1/1). This give me the probability of 1/18 as well.

So I said I would like to be the 5th person for another reason. Last person has an unlikely chance that the people before him would not pick the blue chip (1/18). The 1st person has a (1/18) shot in the dark. While there is a 7/9 chance that the other people will pick the wrong chip. Leaving him a 1/14 chance of picking the blue chip.

I will pick up on part b later if the first part is correct. Does what I did make sense?

For both parts (a) and (b), I get that P{blue} is independent of your position. Hint: forget combinatorial formulas: look instead at the sample space, and what the event {blue in position i} looks like. Note: this assumes that when you pick a chip you have absolutely no information about colors already chosen by others, if any. Everything changes if you do have some information before picking.

RGV

 

[Hodgey8806]

I can see that if information is unknown, then position doesn't matter. But doesn't that change a bit for b?

I can see that for person 1, the probability is 2/18 = 1/9

But for the 5th person, I still used a combinatorial formula: [(2c0)(16c4)/(18c4)]*2/14 + [(2c1)(16C3)/(18c4)]*(1/14)] = 13/153 + 4/153 = 1/9

And then the last person has the probability [(2c1)(16c16)/(18c17)]*1 = 1/9 .

So again, I'm seeing that the probability is independent of position and each one still has an equal chance. Is this correct?