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Binomial distribution 
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#1
Feb2912, 02:50 AM

P: 1,005

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} * (10.25)^{5} = K(10,5)*0.25^{5}*0.75^{5} Or more generally for any binomially distributed outcome: 1) p(x=r) = p^{r}*(1p)^{nr}*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? 


#2
Feb2912, 07:55 AM

P: 199

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=1p=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. 


#3
Feb2912, 01:31 PM

P: 1,005

hmm...
you were to suppose that each side was equally probable :) 


#4
Feb2912, 02:10 PM

P: 199

Binomial distribution
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? 


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