Question

According to the Institute of Health, 32% of women and 17% of men will suffer hip fractures due to falls. Find the probability in a random sample of 5 women, exactly 3 have suffered a broken hip from falls?

Answer 1

`C_(5,3)(.32)^3(.68)^2~~0.3`

Explanation:

We can use a binomial probability to find this. The general form is:

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

Here we have the probability of a woman suffering a broken hip is 32% `(p=0.32)`, meaning the probability a woman won't break a hip is `(~p=0.68)`. We're looking at 5 women `(n=5)` and we want to see the probability of having exactly 3 women who have broken a hip `(k=3)`:

`C_(5,3)(.32)^3(.68)^2=20(0.032768)(0.4624)~~0.3=30%`