Question

Records at a particular hospital show that 11 percent of patients are admitted for surgery, 14 percent for obstetrics, and 1.9 percent for both surgery and obstetrics. Suppose that a patient admission record is selected at random. (Enter answers below as decimals.)

(a) The probability that the selected patient admission record was not for surgery is  .

(b) The probability that the selected patient admission record was for surgery or obstetrics (or both) is  .

(c) If the selected patient admission record was for surgery, what is the probability that it was also for obstetrics?  .

Answer 1

A: Event of a patient admission record was for Surgery

P(A) = 11/100 = 0.11

B : Event of a patient admission record was for obstetrics

P(B) = 14/100

P(A and B) = 1.9/100 = 0.019

(a) Probability that the selected patient admission record was not for surgery is = P(Not A) = 1-P(A) = 1-0.11 = 0.89

Probability that the selected patient admission record was not for surgery is = 0.89

(b) The probability that the selected patient admission record was for surgery or obstetrics (or both) is = P(A or B)

By addition theorem,

P(A or B) = P(A) + P(B) - P(A or B) = 0.11+0.14-0.019=0.231

The probability that the selected patient admission record was for surgery or obstetrics (or both) is = 0.231

(c) If the selected patient admission record was for surgery, what is the probability that it was also for obstetrics = P(B|A)

By Multiplication theorem of probability,

P(A and B) = P(A) P(B|A)

If the selected patient admission record was for surgery, what is the probability that it was also for obstetrics = 0.1727