Thread: chance of picking colors in pairs.

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

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

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

Therefore the probability is
$\displaystyle
\frac{1}{\frac{11!}{5!6!}}
$

which comes down to

$\displaystyle \frac{1}{462}$

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

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

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

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 $\displaystyle \frac{10}{22}\times \frac{12}{21}$, 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 $\displaystyle \frac{2.12.10}{22.21}$.

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 $\displaystyle \frac{2.11.9}{20.19}$.

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 $\displaystyle \frac{7.6}{12.11}$.

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

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

Grandad

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

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

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 $\displaystyle \frac{10}{22}\times \frac{12}{21}$, 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 $\displaystyle \frac{2.12.10}{22.21}$.

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 $\displaystyle \frac{2.11.9}{20.19}$.

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 $\displaystyle \frac{7.6}{12.11}$.

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

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

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.

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

Hi Grandad,

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