Question

The survival rate during a risky operation for patients with no other hope of survival is 90%. What is the probability that exactly four of the next five patients survive this operation? (Give your answer correct to three decimal places.)

Answer 1

The survival rate during a risky operation for patients with no other hope of survival is 90%.

Therefore the probability of survival for these patients during the risky operation = 90/100 =0.9

Let us the denote the event of a patient surviving the operation as A and not surviving as B.

Hence,

P(A) = 0.9   &    P(B) = 1 - P(A) = 1 - 0.9 = 0.1

Q) What is the probability that exactly four of the next five patients survive this operation?

So, exactly four of the next five patients must survive this operation which can be possible through any of the below possible sequences.

Also, A and B are independent events and for independent events A & B,  

1)  (which means the first four patients survive the last doesn't)

= P(A)*P(A)*P(A)*P(A)*P(B) = 0.9*0.9*0.9*0.9*0.1 = 0.06561

2)   (which means second last patient doesn't survive, rest all survive)

= P(A)*P(A)*P(A)*P(B)*P(A) = 0.9*0.9*0.9*0.1*0.9 = 0.06561

3) (which means 3rd patient dies, rest all survive)

= P(A)*P(A)*P(B)*P(A)*P(A) = 0.9*0.9*0.1*0.9*0.9 = 0.06561

4) (which means 2nd patient dies, rest all survive)

= P(A)*P(B)*P(A)*P(A)*P(A) = 0.9*0.1*0.9*0.9*0.9 = 0.06561

5) (which means first patient dies, rest all survive)

=  P(B)*P(A)*P(A)*P(A)*P(A) = 0.1*0.9*0.9*0.9*0.9 = 0.06561

Probability that exactly four of the next five patients survive this operation =

Sum of above 5 probabilities = 0.06561+0.06561+0.06561+0.06561+0.06561 = 5(0.06561) = 0.328