A box contain 100 balls of which r are red. Suppose that the balls are drawn from the box one at a time at random, without replecement. Determine:

A. the probability that the fiftieth ball drawn will be red?

B. the probability that the last ball drawn will be red?

As long as we don't know the results of the previous draws, the probability remains the same.

r/100

In how many ways can we arrange the non-Red balls?

We have to select (100-r) postions out of 100

Answer: 100P(100-r) = 100!/r! (P means Permutations) (nPr = n!/(n-r)! )

In how many ways the (remaining) red balls : 1

Total number of ways: X = 100!/r!

A. One red ball has been placed in 50th position.

Inhow many ways can we arrange the non-red balls? We have 100-r balls to place in 99 positions

Answer : Y = 99P(100-r) = 99!/[99 - (100-r)]!

= 99!/(r -1)!

Remaining red balls can be positioned in only 1 way


Probability = Y/X = [99!/(r -1)!]/[(100)! /r!]

= 99! r!/100!(r -1)!

= r/100

B. Same answer as A.

Let me try to be a little more concise in how I arrived at my conclusion.

a) If we had been told "N red balls been selected in the first 49 picks", we could use that information to modify the probability of the 50th ball. We have not been given that information. The probability is therefore as if this was the first ball selected, that being r/100.

b) Same logic.


Stated another way:

Without replacement can only happen if you have learned what has not been replaced.

In Bingo, if I have selected 25 numbers already, the probablity the next number will be B5 is 1/50 (assuming it wasn't one of the first 25 numbers, in which case the probablity is 0).

If however I secretly draw 25 numbers, the probablity that the 26th number is B5 is still 1/75. Since the first 25 numbers are unknown we can't apply that knowledge to adjust the odds of the current ball.

Very good, Polluxcastor. Makes sense but not perfect mathematical sense!

Hence I would stick to my laboured solution!

I wonder whether you (now or in the past) have actually verified your intuitive correct result Mathematically!

Problem with your logic that I have is:

I take out one ball: probability of red = r/100

Next (without replacing the first) ball: is one out of 99 and no. of reds could be r or r -1.

So I would not be able to say (without calculations) that probability is still r/100.

Unless of course I put the first ball back!

Let's take this scenario, we have out box with 100 marbles, 10 (for now r has been given a value) of them are red. We have 100 cups numbered 1 - 100.

Now we take a marbles out, one at a time and put one under each cup.

a) What is the probability there is a red marble under cup 50.

b) What is the probability there is a red marble under cup 100.

Now lets modify the scenario. We reveal the marbles as we pick them.

Given that 9 of the first 49 marbles are red, what is the probability that marble 50 is red? It is now (10 - 9) / (100 - 50) or 1/50. It has changed because revealing the color removed the marble from the probabilities. It is the action of revealing, not drawing that changes the probabilities.

If we shuffle a deck of cards and deal them to 4 people face down. What is the expectation each has of the number of spades? 3.25

Now each person picks up their cards and looks at them.

Person 1 sees 10 spades, and thinks "there are 3 spades in the other 3 hands"
Person 2 sees 0 spades, and thinks "there are 13 spades in the other 3 hands"
Person 3 sees 1 spade, and thinks "there are 12 spades in the other 3 hands"

Now I have been walking around the table and saw what 1, 2 and 3 have. I now know that person 4 has 2 spades. But no one else but person 4 knows that. Again, it is the action of revealing that solidifies a probability.

I hope that cleared things up a little. (I hope I haven't hijacked the thread)

PS. Lets do it mathematically, r red marbles out of 100. Draw one w/o revealing the color. We have reduced the number of marbles to 99 and the number of red to (r-r/100). So the probability the next marble picked will be red is:
(r-r/100)/99
r(1-1/100)/99
r*.99/99
r/100

Hi PolluxCastor:

Thanks for taking the time.

One other thought: I wonder whether your first (intuitive) answer be considered acceptable in an examination or test?

Jiten