Question

in a study of helicopter usage and patient survival, among the 47,032 patients transported by helicopter, 207 of them left the treatment center against medical advice, and the other 46,825 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
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the total number of patients and non - AMA patients

The total number of patients transported by helicopter is $n = 47032$, and the number of patients who did not leave against medical advice (AMA) is $m=46825$.

Step2: Use the hypergeometric probability formula for non - replacement

The hypergeometric probability formula for choosing $k = 60$ non - AMA patients out of a sample of $N = 60$ patients (where the population size is $N_{total}=47032$ and the number of non - AMA patients in the population is $K = 46825$) is $P=\frac{\binom{K}{k}\binom{N_{total}-K}{N - k}}{\binom{N_{total}}{N}}$. In this case, since we want to choose 60 non - AMA patients out of 60 selected patients, the formula simplifies to $\frac{\binom{46825}{60}\binom{47032 - 46825}{60-60}}{\binom{47032}{60}}=\frac{\binom{46825}{60}}{\binom{47032}{60}}$.
We can also think of it in a sequential - probability way. The probability that the first patient selected is not AMA is $\frac{46825}{47032}$. The probability that the second patient selected is not AMA given that the first one was not AMA is $\frac{46824}{47031}$, and so on.
The probability that none of the 60 patients left against medical advice is:
\[P=\prod_{i = 0}^{59}\frac{46825 - i}{47032 - i}\]
\[P=\frac{46825}{47032}\times\frac{46824}{47031}\times\cdots\times\frac{46825 - 59}{47032-59}\]
\[P\approx0.779\]

Answer:

$0.779$