Posted by Vikki on Friday, February 24, 2012 at 12:06am.
Suppose that an unfair coin comes up heads 54.1% of the time. The coin is flipped a total of 13 times.
a) What is the probability that you get exactly 6 heads?
b) What is the probability that you get exactly 6 tails?
c) What is the probability that you get at most 9 tails?

Probability  MathMate, Friday, February 24, 2012 at 1:31pm
We would use the binomial expansion of
(0.541+0.459)^13, where n=13.
The probability of r occurrences of head is given by (n,r)(0.541)^r(0.459)^(nr), where (n,r) stands for "n choose r" or
(n,r)=n!/(r!(nr)!).
(a) For 6 heads, r=6, n=13
P(6H)=(13,6)0.541^6*0.459^7
=1716*0.02507*.004292
=0.1847
(b) exactly 6 tails is the same as exactly 7 heads, so calculate P(7H) using the above formula.
(c) at most 9 tails means at least 4 heads.
P(≤9T)
=P(≥4H)
=P'(<4H)
By Kolmogorov's second axiom, we can write
=1P(<4H)
=1(P(0H)+P(1H)+P(2H)+P(3H))
Note: You can view Kolmogorov's second axiom at:
http://mathworld.wolfram.com/KolmogorovsAxioms.html