Question

In a study of helicopter usage and patient​ survival, among the

42 comma 75442,754

patients transported by​ helicopter,

194194

of them left the treatment center against medical​ advice, and the other

42 comma 56042,560

did not leave against medical advice. If

5050

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

nothing.

Answer 1

Given:

Total number of patients = 42754

Number of people who did not leave the treatment center against medical advice = 42560

Probability, p = Number of favourable events/Total number of events

= 42560/42754

= 0.9955

p = 0.9955

If 50 of the subjects transported by helicopter are randomly selected without​replacement.

So n = 50

The probability that none of 50 left the treatment center against medical​ advice :

Probability = (50C50) * (0.9955)^50 * (1-0.9955)^50-50

= 1 * (0.9955)^50 * 1

= (0.9955)^50

= 0.7981

Therefore the probability that none of them left the treatment center against medical​ advice is 0.7981