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)^(n-r), where (n,r) stands for "n choose r" or

(n,r)=n!/(r!(n-r)!).

(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

=1-P(<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

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