if the woman is 85, then the probability of having a hip fracture is .38
this means that the probability of not having a hip fracture is 1 - .38 = .62
p(having) + p(not having) = .38 + .62 = 1.0 as it should.
since this is a binomial distribution, you have to use the binomial distribution formula to get the answers.
that formula is:
p(x) = C(n,x) * p^x * q^(n-x)
n is equal 6 because you're dealing with 6 women.
x is equal to the number of hip injuries among the 6 women.
p is equal to the probability of having a hip injury.
q is equal to the probability of not having a hip injury.
C(n,x) is the combination formula of n! / (x! * (n-x)!)
as an example:
when x is equal to 0, the formula tells you that the probability is:
p(0) = C(6,0) * .38^0 * .62^6
when x is equal to 1, the formula tells you that the probability is:
p(1) = C(6,1) * .38^1 * .62^5
when x is equal to 2, the formula tells you that the probability is:
p(2) = C(6,2) * .38^2 * .62^4
etc down to x is equal to 6, the formula tells you that the probability is:
p(6) = C(6,6) * .38^6 * .62^0
you can't solve the problem without using the formula.
when you apply the formula, you get the following:
x (n-x) C(6,x) p^x q^(n-x) p^x*q^(n-x) C(6,x)*p^x*q^(n-x)
0 6 1 1 0.056800236 0.056800236 0.056800236
1 5 6 0.38 0.091613283 0.034813048 0.208878286
2 4 15 0.1444 0.14776336 0.021337029 0.320055438
3 3 20 0.054872 0.238328 0.013077534 0.26155068
4 2 15 0.02085136 0.3844 0.008015263 0.120228942
5 1 6 0.007923517 0.62 0.00491258 0.029475482
6 0 1 0.003010936 1 0.003010936 0.003010936
sum of all probabilities equals >>>>>>>>> 1
you now have all the information you need to solve the problem.
the problem states:
IF SIX WOMEN AGED 85 ARE RANDOMLY SELECTED WHAT IS THE PROBABILITY THAT FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
if fewer than 2 of them suffer, then that would be equal to the probability of p(0) + p(1).
from the chart, p(0) = 0.056800236 and p(1) = 0.208878286
add these together and you get the probability that fewer than 2 will suffer is equal to .265678522.
you only need to calculate the probability of 0 or 1 sufferers because the problem stated that you wanted the probability that fewer than 2 suffered.
i shows you the whole table to show you how the formula works and to demonstrate that the total probability had to be equal to 1 as it was based on the use of those formulas.
in the table, that last column on the right is equal to p(x).
example:
when x = 0, p(x) = 0.056800236
when x = 6, p(6) = 0.003010936
