Question

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

57 comma 99957,999

patients transported by helicopter,

292292

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

57 comma 70757,707

did not leave against medical advice. If

6060

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?

Answer 1

Out of 57,999 patients, 57,707 patients did not leave the

treatment center against medical advice.

The probability that a patient will not leave against medical

advice = 57707 / 57999 = 0.9950.

The probability that a patient will leave against medical advice

= 1 - 0.9950 = 0.0050.

Let X be the random variable denoting the number of subjects

who leave against medical advice out of 60 subjects.

Hence, X ~ Bin(60, 0.0050).

The probability that none of the 60 subjects leave the

treatment center against medical advice = P(X = 0)

= 0.7403. (Ans).