chance of picking colors in pairs.

**Dindia** · Dec 22nd 2008, 08:00 AM

Hi All!

I have 10 black balls and 12 white balls (mixed).
The balls are randomly sorted to pairs.

Q: what is the chance of picking 5 pairs with white and black balls in each and 1 pair of 2 remaining white balls?

I don't have a clue how to begin solving it.
Hope you can help.

Thank you so much in advance!
Danny

**Pim** · Dec 24th 2008, 02:08 AM

Well, I'm not entirely sure, but I do have an idea how it could be done.

What you do is fix the first ball of each pair, then "move" the other ones
(Drawing with O as black and X as white)

|O|O|O|O|O|X|X|X|X|X|X|
| | | | | | | | | | | |

Now, the leftover balls can be sorted in

$<br /> \frac{11!}{5!6!}<br />$ ways.
only 1 of these is correct. (Namely with first all black balls then all white ones.)

Therefore the probability is
$<br /> \frac{1}{\frac{11!}{5!6!}}<br />$

which comes down to

$\frac{1}{462}$

which seems about right to me. (Again, I'm not 100% positive this is the correct way, but the outcome seems to be right.)

**Grandad** · Dec 24th 2008, 08:02 AM

Hello Danny

Originally Posted by Dindia

Hi All!

I have 10 black balls and 12 white balls (mixed).
The balls are randomly sorted to pairs.

Q: what is the chance of picking 5 pairs with white and black balls in each and 1 pair of 2 remaining white balls?

I don't have a clue how to begin solving it.
Hope you can help.

Thank you so much in advance!
Danny

I have to choose a total of 12 balls from the 22 original balls, one at a time, without replacing each choice.

The probability that my first choice is a white ball is $\frac{12}{22}$ . If this ball is white, the probability that my second choice is black is $\frac{10}{21}$ .

So the probability that I choose white followed by black is $\frac{12}{22}\times\frac{10}{21}$ .

But we want the first pair to be one of each colour. So we must also allow the choice to be black followed by white. This is $\frac{10}{22}\times \frac{12}{21}$ , by a similar argument. You'll see that this is the same fraction as before. So the probability that the first pair I choose contains a white ball and a black in either order is $\frac{2.12.10}{22.21}$ .

If the first pair does contain one of each colour, then there are 11 white and 9 black remaining. So, in the same way as before, the probability that the second pair I choose contains a ball of each colour is $\frac{2.11.9}{20.19}$ .

Continue in the same way for the 3rd, 4th and 5th pairs.

The sixth pair is different. By now, if all the previous events have occurred, there are 7 white and 5 black balls left. The probability that the next two chosen are both white is $\frac{7.6}{12.11}$ .

Multiply all these six probabilities together to obtain the overall probability of all these events occurring. Expressed as factorials, this comes to $\frac{2^5.12!.(10!)^2}{(5!)^2.22!}$ .

Or, if you're careful you can cancel the numbers to get an exact answer, which is $\frac{576}{46,189} \approx 0.0125$ .

Grandad

**awkward** · Dec 25th 2008, 05:11 PM

Originally Posted by Grandad

Hello Danny

I have to choose a total of 12 balls from the 22 original balls, one at a time, without replacing each choice.

The probability that my first choice is a white ball is $\frac{12}{22}$ . If this ball is white, the probability that my second choice is black is $\frac{10}{21}$ .

So the probability that I choose white followed by black is $\frac{12}{22}\times\frac{10}{21}$ .

But we want the first pair to be one of each colour. So we must also allow the choice to be black followed by white. This is $\frac{10}{22}\times \frac{12}{21}$ , by a similar argument. You'll see that this is the same fraction as before. So the probability that the first pair I choose contains a white ball and a black in either order is $\frac{2.12.10}{22.21}$ .

If the first pair does contain one of each colour, then there are 11 white and 9 black remaining. So, in the same way as before, the probability that the second pair I choose contains a ball of each colour is $\frac{2.11.9}{20.19}$ .

Continue in the same way for the 3rd, 4th and 5th pairs.

The sixth pair is different. By now, if all the previous events have occurred, there are 7 white and 5 black balls left. The probability that the next two chosen are both white is $\frac{7.6}{12.11}$ .

Multiply all these six probabilities together to obtain the overall probability of all these events occurring. Expressed as factorials, this comes to $\frac{2^5.12!.(10!)^2}{(5!)^2.22!}$ .

Or, if you're careful you can cancel the numbers to get an exact answer, which is $\frac{576}{46,189} \approx 0.0125$ .

Grandad

Hi Grandad,

I agree with your analysis up to a point, but it assumes the white-white pair is the last pair taken. I think the problem statement, on the other hand, allows it to be any one of the 6 pairs. Each of these 6 outcomes has the same probability, so I think we need to multiply your answer by 6, for a final probability of 0.074823.

**Grandad** · Dec 25th 2008, 09:21 PM

Hi Danny

Originally Posted by awkward

Hi Grandad,

I agree with your analysis up to a point, but it assumes the white-white pair is the last pair taken. I think the problem statement, on the other hand, allows it to be any one of the 6 pairs. Each of these 6 outcomes has the same probability, so I think we need to multiply your answer by 6, for a final probability of 0.074823.

I think you could well be right. I certainly assumed from the wording of the question - especially from the use of the word 'remaining' - that the white-white pair was the last pair to be revealed. But if it can come anywhere in the order, then you're right to multiply by 6.

Grandad

**awkward** · Dec 26th 2008, 01:29 PM

Hi Grandad,

The problem statement is, I concede, unclear or at least ... awkward.

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