I thought that someone probably could help me with this probability question (such pun many funny wow)

I am  assuming that each colour is equally represented so each spin there is 1/3 chance of getting a yellow.

$$\mbox{P(7out of 10 yellow) =}10C7 * (1/3)^7 * (2/3)^3$$

$${\left({\frac{{\mathtt{10}}{!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}^{{\mathtt{7}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{3}}} = {\mathtt{0.016\: \!257\: \!684\: \!296\: \!093\: \!1}}$$

We want the wheel to come up "yellow" 7 out of 10 times....so we have  C(10,7)

And the number of possible outcomes is just  3^10

So we have  C(10,7) / 3^10  ≈ .002   ≈ 2/1000   ≈ 1/500

Sorry Phill, but I think that your answer is wrong.

The number of successful outcomes should be 8*C(10,7), not simply C(10,7).

Consider a sequence consisting of three non-yellows followed by seven yellows, the number of ways in which that could occur is 2*2*2*(7 1's) = 8. That is, red or blue, red or blue, red or blue, yellow, yellow, ... , yellow. The numbers will be the same (but in different orders) for each successful sequence.

That gives you a probability of 8*C(10,7)/3^10.

The other way of looking at the problem is through probabilities. Consider the same sequence as above, three non-yellows followed by seven yellows. The probability of that happening is ((2/3)^3)*((1/3)^7) = 8/(3^10). All successful sequences will have the same probability so the total probability will be (8/(3^10))*C(10,7).

Thanks, Bertie...I see where I made my mistake !!!!