binomial distribution


by aaaa202
Tags: binomial, distribution
aaaa202
aaaa202 is offline
#1
Feb29-12, 02:50 AM
P: 995
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)*0.255*0.755
Or more generally for any binomially distributed outcome:

1) p(x=r) = pr*(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)/4n
How do you show that 1) and 2) are equivalent?
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alan2
alan2 is offline
#2
Feb29-12, 07:55 AM
P: 188
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.
aaaa202
aaaa202 is offline
#3
Feb29-12, 01:31 PM
P: 995
hmm...

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

alan2
alan2 is offline
#4
Feb29-12, 02:10 PM
P: 188

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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