Repeated and conditional probability question

1. Aug 11, 2015

erisedk

1. The problem statement, all variables and given/known data
Consider a bag containing 10 balls of which a few are black balls. Probability that bag contains exactly 3 black balls is 0.6 and probability of bag containing exactly 1 black ball is 0.4. Now, balls are drawn from the bag, one at a time, without replacement, till all black balls have been drawn. The probability that this procedure would end at the 6th draw is p

2. Relevant equations

3. The attempt at a solution
If bag contains 3 black balls, P(6) = 0.6 * (3.2.7.6.5.1)/(10.9.8.7.6.5) * 5! = 0.6 (5! because that's the possible ways of arranging 3, 2, 7, 6, 5 in different orders)

If bag contains 1 black ball, P(6) = 0.4 * (9.8.7.6.5.1)/(10.9.8.7.6.5) = 1/25

So, P(6) = 0.6 + 1/25 = 16/25

Which is wrong.

2. Aug 11, 2015

PeroK

I'm not sure why you are multiplying by 5!

Why not think about it from the last ball, imagining that all 10 balls are picked?

3. Aug 11, 2015

Ray Vickson

If there are $B = 1$ black balls present, the conditional probability of getting it on draw 6 is
$$P(6|B=1) = \frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{1}{5} = \frac{1}{10},$$
as you have more-or-less indicated.

If there are $B = 3$ black and $O = 7$ other colors, the conditional probability you want can be obtained from the probability $P_{2,5}$ of drawing exactly $k =2$ black in the first $n = 5$ draws, then following that with a draw of the black from the remaining five balls. You can get $P_{2,5}$ from the hypergeometric distribution; see, eg.,
https://en.wikipedia.org/wiki/Hypergeometric_distribution .

4. Aug 11, 2015

RUber

I like your logic. But, given that it is wrong, I will suggest another approach.
This may be more brute force than some more elegant method, but it should work:
If the bag has three black balls, the ways you might choose 2 of the 3 within the first five draws is 3c2*5c2, multiplied by the ways you could draw the black balls 7c3 divided by the total ways to pull 5 balls = 10c5, multiplied by the associated probability of the draw pattern which you have above.
This gives:$(.6) * \frac{ 7*6*5*3*2*1}{10*9*8*7*6*5}\frac{ 3!}{2}\frac{ 5!}{3!*2}\frac{ 7!}{4!3!}\frac{ 5!}{3!*2}\frac{ 5!5!}{10!}$

If the bag has one black ball, your number is correct.

5. Aug 11, 2015

PeroK

With one black ball, it could be drawn anywhere between 1 and 10 with equal likelihood, so the probability is 1/10 that it's the 6th.

With 3 black balls, if the 6th is the last black, then that's equivalent to the last 4 being white and the 6th being black. If you imagine the balls being drawn in reverse order.

6. Aug 12, 2015

Ray Vickson

Along these same lines: number the balls 1--10 and look at permutations. There are 7 choices for position 10, then (given the previous choice), 6 choices for position 9, then 5 for position 8 then 4 for position 7. Then there are three choices for position 6 (one of the unused black balls). After that, the positions 1--5 may be filled in $5!$ different ways. So, the total number of permutations having the last 4 places non-black and position 6 black is $7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 5!$, so the desired (conditional) probability is
$$P(6|B=3) = \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 5!}{10!} = \frac{1}{12}$$
This same result is obtained from the hypergeometric approach and the others suggested.

7. Aug 12, 2015

PeroK

One can also start from the 6th ball and look at the probability that the 6th is black and the last 4 are white. Hence, the following have equal probability:

A) the 6th ball is the last black.

B) the first 4 are white and the 5th is black.

C) the first is black and the next 4 are white.

8. Aug 12, 2015

D H

Staff Emeritus
When you are solving a probability problem (or any technical problem, for that matter), it helps to step back and ask yourself whether the result you just obtained makes any sense.

The result you arrived at doesn't make any sense.

What your result says is that if the bag does indeed contain 3 black balls, the procedure must end at the 6th draw. That of course is wrong; you could be lucky and draw three black balls in a row from the start, or be unlucky and not draw the third black ball until the 10th draw. All values between 3 and 10 are possible.

9. Aug 12, 2015

Ray Vickson

The question was to find the probability that the last black ball is drawn at draw 6. Of course, the last black can occur in any draw from 3 to 10, but the question was specifically about it happening at draw 6.

That being said, I cannot make any sense of the expression the OP wrote, but I don't know what the basis of his error may be.

10. Aug 12, 2015

D H

Staff Emeritus
I realize what the question is asking. I wrote about the calculation
$$P(\text{ends on draw 6}) = P(\text{3 balls}) P(\text{ends on draw 6} | \text{3 balls}) + P(\text{1 ball}) P(\text{ends on draw 6} | \text{1 ball})$$

which is correct, but the OP then calculated $P(\text{ends on draw 6} | \text{3 balls})$ as identically one.

11. Aug 12, 2015

Ray Vickson

Yes, I guess he did that, but in a long-winded way.