Question

In a study of helicopter usage and patient survival, among the 56,074 patients transported by helicopter, 235 of them
left the treatment center against medical advice, and the other 55,839 did not leave against medical advice. If 40 of
the subjects transported by helicopter are randomly selected without replacement, what is the probability that none
of them left the treatment center against medical advice?
The probability is
(Round to three decimal places as needed.)

Answer 1

Using the hypergeometric distribution, it is found that the probability is 0.845 = 84.5%.

--------------------

• The patients are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

The parameters are:

• x is the number of successes.
• N is the size of the population.
• n is the size of the sample.
• k is the total number of desired outcomes.

Combination formula:

is the number of different combinations of x objects from a set of n elements, given as:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

--------------------

• 56074 patients, thus $N = 56074$
• Sample of 40, thus, $n = 40$
• 235 left against medical advice, thus, $k = 235$.

The probability that none left against medical advice is P(X = 0), so:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,56074,40,235) = \frac{C_{235,0}*C_{55839,40}}{C_{56074,40}} = 0.845$

The probability is 0.845 = 84.5%.

A similar problem is given at https://brainly.com/question/24008577