Question

A coin is tossed 8 times. Find the probability of getting at most 5 heads?

Answer 1

`1-C_(8,6)(.5)^6(.5)^(2)-C_(8,7)(.5)^7(.5)^(1)-C_(8,8)(.5)^6(.5)^(0)~~0.8555`

Explanation:

We can do this using binomial probability. The relation I like to start with is:

`sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(n-k)=1`

Here we have `n=8, p=.5` and we're looking for the probability for `0<=k<=5`. We can do this either by adding up the 6 terms within 0 through 5, or we can subtract from 1 the terms for `5 < k<=8`. Let's do it that second way:

`1-C_(8,6)(.5)^6(.5)^(2)-C_(8,7)(.5)^7(.5)^(1)-C_(8,8)(.5)^6(.5)^(0)~~0.8555`