Question 1206328
this looks like a binomial distribution type problem.

formula for that is:


p(x) = p^x * q^(n-x) * c(n,x)


c(n,x) = n! / (x! * (n-x)!)


in this problem:
n = 13
x = 0 to 13
p = probability of heads = .5
q = probability of tails = .5


the total probability will be equal to 1.


the excel spreadsheet shown below gives you all the probabilities.


<img src = "http://theo.x10hosting.com/2024/030101.jpg">


your solutions are:


sum of all probabilities equals 1.


probability of getting exactly 3 heads equals 0.034912109.


probability of getting at least 2 heads = 0.998291016.


probability of getting at most 9 heads = 0.953857422.


as an example of the calculations involved, we'll look at the probability of getting at least 2 heads.


since the probability of getting at least 2 heads is the same as 1 minus the probability of getting less than 2 heads, and it's much easier to make two calculations rather than 12 calculations, we'll calculate 1 minus the probability of getting less than 2 heads..


formula is p(x) = p^x * q^(n-x) * c(n,x).
c(n,x) = n! / (x! * (n-x)!)


n = 13
p = .5
q = .5


p(x = 0) = .5^0 * .5^13 * c(13,0) = .0001220703125.
p(x = 1) = .5^1 * .5^12 * c(13,1) = .0015869141.


p(x = 0 to 1) = .0017089844.
1 minus that = .9982910156.
round that to 9 decimal digits (same as excel), and you get .998291016.
that's the same as excel provides you.


the solution made use of the fact that p(x >= 2) is the same as 1 minus p(x < 2).
this allowed many less manual calculations to be performed.