Probability; 100 balls, r are red.

[1MileCrash]

One ball is selected from 100 at a time, without replacement. Of these balls, r of them are red.

What is the probability that the first ball picked will be red?

The 50th?

The last?

Solution:

I don't understand why it matters which pick it is. I know that "without replacement" changes things quite a bit, but in my mind if the picks are random then that's like arranging them in a random line and having them being picked in that order, in which case the probability that the 50th is red is the same as that of the first or last. So my intuition says r/100 for everything, but none of the previous section in the book has an applicable equation to confirm my intuition. Is my thinking correct?

 

[Punkyc7]

What if there were three red balls out of the 100 balls and the first 3 you selected were red. Would the probability that the 50th ball is red be 3/100?

 

[1MileCrash]

What?

 

[1MileCrash]

The probability of me picking a red ball on the 50th ball after already picking all 3 red balls is zero, but the probability of that happening in the first place (which you didn't consider) is no different than the 50th, 51st, and 52nd being red before the picking started. So I have no idea what you're trying to say?

 

[Ray Vickson]

One ball is selected from 100 at a time, without replacement. Of these balls, r of them are red.

What is the probability that the first ball picked will be red?

The 50th?

The last?

Solution:

I don't understand why it matters which pick it is. I know that "without replacement" changes things quite a bit, but in my mind if the picks are random then that's like arranging them in a random line and having them being picked in that order, in which case the probability that the 50th is red is the same as that of the first or last. So my intuition says r/100 for everything, but none of the previous section in the book has an applicable equation to confirm my intuition. Is my thinking correct?

As long as on the 50th or 100th pick you have _no_ information about the results of previous picks, then with or without replacement, it makes no difference: the probability of red on pick #1 is the same as on pick # 50 and on pick # 100.

How can you convince yourself of this? Well, a pick sequence = a random permutation of the numbers from 1 to 100. The number of different permutations having a red ball in position 1 is the same as the number of permutations having red in position 50 or in position 100.

RGV

 

[1MileCrash]

Thanks, that makes perfect sense.

In what types of situations will with/without replacement make a difference?

 

[Ray Vickson]

Thanks, that makes perfect sense.

In what types of situations will with/without replacement make a difference?

It will matter if we are asking for the probability of drawing a certain number of red balls in a sample of size n > 1 (where 'certain number' includes 0). For example, if r = 10 and we draw 4 balls, the probability of drawing 0 red balls is P0 = (90/100)(89/99)(88/98)(87/97) ≈ 0.6516505491 without replacement and is P0a = (90/100)^4 = 0.6561 with replacent. The differences get more serious for larger sample sizes. For example, if we draw n = 50 items (and r = 10 still) the probability of getting 6 red balls is P6 = C(10,6)*C(90,44)/C(100,50) ≈ 0.2114132170 without replacement and is P6a = 0.1541038341 without replacement.

RGV