Question 982727
 
Given:
An unfair die with heads 56.9% of the time.
Find probability that up to 25 heads show up in 27 throws.
 
Solution:
Binomial probability distribution,
p=0.569
n=27
r=25
So, 
{{{P(n,r)=C(n,r)*p^r*(1-p)^(n-r)}}}
 
{{{P(27,26)=C(27,26)*0.569^26*(1-0.569)^1=4.9976*10^-6}}}
{{{P(27,27)=C(27,27)*0.569^27*(1-0.569)^0=2.444*10^-7}}}
 
Therefore probability for getting up to 25 heads
P(27,0-25)=1-P(27,>25)=1-P(27,26)-P(27,27)
={{{1-4.9976*10^(-6)-2.444*10^(-7)=1-5.242*10^(-6)=0.9999948}}}