Suppose you have a coin with 4 fair sides, flip it 5 times, and want to know the probability of 5 heads. This is K(10,5) * (0.25)^{5} * (1-0.25)^{5} = K(10,5)*0.25^{5}*0.75^{5} Or more generally for any binomially distributed outcome: 1) p(x=r) = p^{r}*(1-p)^{n-r}*K(n,r) But also we must have that: 2) p(x=r) = K(n,r)/total combinations = K(n,r)/4^{n} How do you show that 1) and 2) are equivalent?
They're not equivalent. Your concept in 2) is not applicable in this case because the events are not equally likely. Take for example a single coin flip of a fair coin where p=1-p=1/2. The probability of heads is the number of events resulting in heads divided by the total number of possible outcomes or 1/2. All events are equally likely. Now consider a weighted coin where p=3/4=p(heads). The number of ways to get heads is still 1 and the total number of possible outcomes is still 2 but p(heads) does not equal 1/2.
Ooops, I'm really sorry, you have a 4 side coin. My bad. Anyway, your expressions are incorrect. I'm assuming only one side of the coin is heads. In that case, the probability of 5 heads is (0.25)^5=(1/4)^5. Alternatively, the number of ways to get 5 heads is 1 and the number of possible outcomes is 4^5, so P(5 heads)=1/(4^5). Same result. Where did the 10 come from?