The probability that a patient recovers from a delicate heart operation is .0.9. What is the probability that exactly 5 of the next 7 patient having this operation survive?
Binomial with n = 7 and p = 0.9. P(X = 5) = 0.124
I dont know the answer and want your help to solve this question
To find the probability that exactly 5 out of the next 7 patients survive the heart operation, we can use the binomial probability formula:
P(X=k) = (nCk) * (p^k) * ((1-p)^(n-k))
Where:
P(X=k) is the probability of having exactly k successes (in this case, 5 patients surviving)
n is the total number of trials or experiments (in this case, 7 patients)
k is the number of desired successes (in this case, 5 patients surviving)
p is the probability of success in each trial (in this case, 0.9 or 90%)
Plugging in the values into the formula:
P(X=5) = (7C5) * (0.9^5) * ((1-0.9)^(7-5))
Calculating the values:
(7C5) = 7! / (5! * (7-5)!) = 21
0.9^5 ≈ 0.59049
(1-0.9)^(7-5) = (1-0.9)^2 = 0.01
P(X=5) = 21 * 0.59049 * 0.01 ≈ 0.1244
Therefore, the probability that exactly 5 out of the next 7 patients survive the heart operation is approximately 0.1244 or 12.44%.
To calculate the probability that exactly 5 of the next 7 patients survive the delicate heart operation, you can use the binomial probability formula. The formula is:
P(X=k) = nCk * p^k * (1-p)^(n-k)
Where:
- P(X=k) represents the probability of exactly k successes (in this case, survival)
- n is the total number of trials (in this case, the number of patients)
- k is the number of successful outcomes (in this case, the number of patients who survive)
- p is the probability of success on a single trial (in this case, the probability that a patient recovers)
- (1-p) is the probability of failure on a single trial (in this case, the probability that a patient doesn't recover)
- nCk represents the number of combinations of n items taken k at a time, which can be calculated as n! / (k!(n-k)!)
Applying this formula to your problem, the values are:
n = 7 (number of patients)
k = 5 (number of patients surviving)
p = 0.9 (probability of a patient recovering)
Now, let's substitute these values into the formula:
P(X=5) = 7C5 * 0.9^5 * (1-0.9)^(7-5)
Calculating the values:
7C5 = (7! / (5!(7-5)!)) = (7! / (5!2!)) = 21
0.9^5 = 0.59049
(1-0.9)^(7-5) = 0.01
Plugging in these values:
P(X=5) = 21 * 0.59049 * 0.01
Calculating further:
P(X=5) ≈ 0.1239
Therefore, the probability that exactly 5 of the next 7 patients having this operation survive is approximately 0.1239 or 12.39%.