SOLUTION: A coin is tossed 13 times.
a) How many different outcomes are possible?
b) What is the probability of getting exactly 3 heads?
c) What is the probability of get
Algebra.Com
Question 1206328: A coin is tossed 13 times.
a) How many different outcomes are possible?
b) What is the probability of getting exactly 3 heads?
c) What is the probability of getting at least 2 heads?
d) What is the probability of getting at most 9 heads?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.
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.
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