# Selecting a Red Ball from an Urn

1. Nov 18, 2007

### e(ho0n3

[SOLVED] Selecting a Red Ball from an Urn

1. The problem statement, all variables and given/known data
An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B and so on. There is no replacement of the balls drawn.)

2. Relevant equations
Axioms and basic theorems of probability.

3. The attempt at a solution
This is similar to the dice problem I had posted about previously here. Let S be the sample space. S = {r, br, bbr, bbbr, bbbbr, bbbbbr, bbbbbbr, bbbbbbbr} where bbbr for example denotes that A drew a black ball, then B drew a black ball, then A drew a black ball, then B drew a red ball. The probability that A selects the red ball is P{r} + P{bbr} + P{bbbbr} + P{bbbbbbr} where P{r} = 3/10, P{bbr} = 7/10 * 6/9 * 3/8, P{bbbbr} = 7/10 * 6/9 * 5/8 * 4/7 * 3/5 and P{bbbbbbr} = 7/10 * 6/9 * 5/8 * 4/7 * 3/6 * 2/5 * 3/4. Is this correct?

2. Nov 18, 2007

### EnumaElish

I think it is.

3. Nov 18, 2007

### e(ho0n3

That's enough closure for me. I'm going to mark this problem as solved. Thank you for looking it over.

4. Nov 28, 2007

### Skins

Classic problem, easy answer. There are 10 balls, 7 red, 3 black. Player A wins by selecting the red ball. Player A goes first. Thus player A can only win on the 1st, 3rd, 5th, or 7th draw. And draws beyond that means there are no more black balls left and only red balls are left. At that point player B would win on the 8th draw,

So,

Let E1 be the event that the 1st red ball is drawn by Player A on the 1st draw
Let E3 be the event that the 1st red ball is drawn by Player A on the 3rd draw
Let E5 be the event that the 1st red ball is drawn by Player A on the 5th draw
Let E7 be the event that the 1st red ball is drawn by Player A on the 7th draw

If E1 occurs (if a red ball is pulled on the first draw) it implies that there are C(9,2) ways that the remaining 2 red balls can be arranged among the 9 remaining balls thus there are C(9,2) ways that event E1 can occur.

Similarly if E3 occurs there are C(7,2) ways the remaining 2 red balls can be arranged among the 7 remaining balls thus there are C(7,2) ways that event E3 can occur.

By similar reasoning there are:

C(5,2) ways that event E5 can occur
C(3,2) ways that event E7 can occur

If we consider the sample space as the set of 3 red and 7 black balls then the number of ways a red ball can be randomly drawn from the set of 10 balls is C(10,3)

Therefore the probability that Player A selects a red ball is given by

P = C(9,2) + C(7,2) + C(5,2) + C(3,2) / C(10,3)

Which computes to ->

P = (36 + 21 + 10 + 3) / 120 = 70/120 = 7/12

Thus the Probability that Player A will draw a red ball is 7/12 or approximately 0.58333...

Skins

5. Nov 29, 2007

### e(ho0n3

Do you see anything wrong with my method?

Your method of counting the ways an event occurs doesn't make sense to me. For example, you wrote that one can arrange two red balls given 9 balls in C(9,2) ways. What do you mean by "arrange"? C(9,2) is the number of ways of picking two balls from a set of 9.

Another example: You wrote that a red ball can be randomly drawn from a set of 10 in C(10,3) ways but C(10,3) is the number of ways of picking three balls from ten.

6. Nov 29, 2007

### Skins

Okay, let me try and word it differently. Consider event E1 which I defined as the event in which Player A gets a red ball on the first draw. Now think in terms of how many ways event E1 can occur. Think of the set of 3 red and 7 black balls as being arranged in a sequence with the first one being red. For example three possible such sequences are

1 2 3 4 5 6 7 8 9 10
R B B B B R B B R B
R R B R B B B B B B
R B B B B B R B B B

The ONLY constraint upon these sequences is that the first ball (representing the first draw by Player A) must be red. Therefore the number of different sequences in which a red ball appears in position 1 (first draw) is determined by the remaining 9 balls two of which are red. Therefore the number of possible sequences is given by C(9,2),

Now, we follow the same line of reasoning for the events E3, E5 and E7 obtaining the values. For example E3 is the event where the first red ball is drawn by Player A on the 3rd draw. In order for this to happen the first two draws by Player A and Player B respectively would have to have been black balls. Therefore if the first red one is pulled on the 3rd draw there will be 7 balls remaining (5 black and 2 red) Therefore the number of ways that the 3rd ball can be the first red ball will be determined by the sequence of the remaining 7 balls given by C(7,2)
With similar reasoning there are C(5,2) and C(3,2) possible sequences that will produce events E5 and E7 respectively.

[/QUOTE]

OK let me clarify. Here we are talking about the number of possible sequences for the entire sample space S of the 10 ( 7 black & 3 red )balls This will be determined by the number of ways the 3 red balls can be arranged among the 10 possible positions of the sequence or C(10,3). In other words C(10,3) represents the size of our sample space.

Since the events E1, E3, E5, and E7 are mutually exclusive (disjoint) we have

P(E1 U E3 U E5 U E7) = P(E1) + P(E3) + P(E5) + P(E7) which is the probability of the first red ball being pulled on the 1st, 3rd, 5th, and 7th draws respectively. From the values we determined above this becomes:

P = C(9/2) / C(10,3) + C(7,2) / C(10,3) + C(5,2) / C(10, 3) + C(3,2) / C(10,3)

or

P = C(9,2) + C(7,2) + C(5,2) + C(3,2) / C(10,3)

These types of counting problems are confusing and often hard to visualize. However, the result I presented here is pretty much the established result according to two textbooks and at least one online source as well as numerous classroom repetitions of this problem.

In any event I hope that this helped clear things up a bit. Sorry if my original explaination was a bit confusing. Think of the problem in terms of the events that will produce a win for player A and how the events can occur and in how many ways.

D.M. Thomas (aka Skins)

Last edited: Nov 29, 2007
7. Nov 29, 2007

### D H

Staff Emeritus
echo1, Skins is correct.

Here's another way to get the same results as Skins.

Suppose that nobody has drawn a red ball up to the nth draw. To get to this point, a black ball had to have been chosen on each of the first n-1 draws, leaving 3 red balls and 7-(n-1) black balls in the urn. The conditional probability of drawing a red ball on the nth draw is thus $\frac 3 {8-n}$. The probability of getting to the nth draw is $\frac 7 {10}\,\frac 6 9\,\cdots \frac {7-(n-2)}{10-(n-2)}$. The total probability of winning on the nth roll is the product of these two. A quick table:
Code (Text):

Draw#  Prior prob.   Win cond. prob.  Win prob.
1         1            3/10           3/10
2        7/10          1/3            7/30
3        7/15          3/8            7/40
4        7/24          3/7            1/8
5        1/6           1/2            1/12
6        1/12          3/5            1/20
7        1/30          3/4            1/40
8        1/120         3/3            1/120
Adding up the probabilities of winning on the odd-numbered draws yields 7/12.

8. Nov 29, 2007

### e(ho0n3

Aha! So you're saying the sample space consists of these 10-character sequences that have three R's and seven B's. I reasoned that the sample space would contain only sequences of the form B...BR because each element of the sample space would represent a possible game where the first person to pick a red ball wins. Why is my reasoning wrong?

I think that should be 3 / (8 - n + 3) since there are 8 - n + 3 black plus red balls remaining.

Last edited: Nov 29, 2007
9. Nov 29, 2007

### D H

Staff Emeritus
Oops. You are correct. I can't blame it on a typo, as there is no slip of the finger that can take one from 8 to 11. Careless typing?? Fortunately, I didn't use that erroneous expression in calculating the probabilities. If you look at my table, it is obvious I used used the correct conditional probability.

10. Nov 29, 2007

### e(ho0n3

Now I know why I got a different answer: I messed up in calculating P{bbbbr}. I wrote that P{bbbbr} = 7/10 * 6/9 * 5/8 * 4/7 * 3/5. The last term should actually be 3/6. Doh!

Anywho, thank you for helping D H and Skins.