- #1
aaaa202
- 1,169
- 2
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?
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?