Question

in a study of helicopter usage and patient survival, among the 57,072 patients transported by helicopter, 271 of them left the treatment center against medical advice, and the other 56,801 did not leave against medical advice. if 60 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 (square).
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the total number of patients

The total number of patients transported by helicopter is $n = 57072$. The number of patients who did not leave against medical - advice is $m=56801$.

Step2: Use the hyper - geometric probability formula for non - replacement

The probability that none of the 60 randomly selected patients left against medical advice is given by the hyper - geometric probability formula. The formula for the probability of $k = 0$ successes (in this case, a success is a patient leaving against medical advice) in a sample of size $r = 60$ from a population of size $N = 57072$ with $K=271$ non - successes (patients who left against medical advice) and $N - K = 56801$ successes (patients who did not leave against medical advice) is:
\[P(X = 0)=\frac{\binom{56801}{60}\binom{271}{0}}{\binom{57072}{60}}\]
Since $\binom{a}{0}=1$ for any non - negative integer $a$, the formula simplifies to:
\[P(X = 0)=\frac{\binom{56801}{60}}{\binom{57072}{60}}\]
\[\binom{n}{r}=\frac{n!}{r!(n - r)!}\]
\[\binom{56801}{60}=\frac{56801!}{60!(56801 - 60)!}\]
\[\binom{57072}{60}=\frac{57072!}{60!(57072 - 60)!}\]
\[P(X = 0)=\frac{\frac{56801!}{60!56741!}}{\frac{57072!}{60!57012!}}=\frac{56801!×57012!}{57072!×56741!}\]
An alternative way is to calculate the probability step - by - step for each selection.
The probability that the first patient did not leave against medical advice is $\frac{56801}{57072}$.
The probability that the second patient did not leave against medical advice given that the first one did not is $\frac{56799}{57071}$ (since one non - leaving patient is already selected and one patient is removed from the total).
For 60 selections, the probability is:
\[P=\frac{56801}{57072}\times\frac{56799}{57071}\times\cdots\times\frac{56801-(60 - 1)}{57072-(60 - 1)}\]
\[P=\prod_{i = 0}^{59}\frac{56801 - i}{57072 - i}\]
Using a calculator or software for the product:
\[P\approx0.741\]

Answer:

$0.741$