The survival rate during a risky operation for patients with no other hope of survival is 87%. What is the probability that exactly four of the next five patients survive this operation? (Give your answer correct to three decimal places.)

I got 3.48

Incorrect

p(s) =0.87
n = 5

P(x =4)
P(x) = nCx p^x(1-p)^n-x

(5C4)(0.87)^4(0.13)^1 = 0.372

Thank you, I left out a complete step is how I missed it...

Sorry for the different names, there is a group of us trying to work these out together and using one computer when we have questions...

No problem

To find the probability that exactly four out of the next five patients survive the operation, you can use the binomial probability formula.

The binomial probability formula is P(x) = (nCx) * p^x * q^(n-x), where:
- P(x) is the probability of getting exactly x successes
- n is the total number of trials or patients
- x is the number of successes
- p is the probability of success on a single trial
- q is the probability of failure on a single trial (q = 1 - p)

In this case, n = 5 (since there are five patients), x = 4 (since we want exactly four patients to survive), and p = 0.87 (since the survival rate is 87%, or 0.87). Thus, q = 1 - p = 1 - 0.87 = 0.13.

Plugging these values into the formula, we get:
P(4) = (5C4) * 0.87^4 * 0.13^(5-4)
= (5! / (4! * (5-4)!)) * 0.87^4 * 0.13
= (5 * 0.87^4 * 0.13)

Calculating this expression, we find:
P(4) = (5 * 0.87^4 * 0.13) ≈ 0.607

Therefore, the probability that exactly four of the next five patients survive this operation is approximately 0.607, or 60.7% when rounded to three decimal places. So your answer of 3.48 is incorrect.