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