A new surgery is successful 80% of the time. If the results of 8 such surgeries are randomly sampled, what is the probability that more than 5 of them are successful?Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.(If necessary, consult a list of formulas.)
Question
A new surgery is successful 80% of the time. If the results of 8 such surgeries are randomly sampled, what is the probability that more than 5 of them are successful?Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.(If necessary, consult a list of formulas.)
Solution
To solve this problem, we can use the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
In this case, we want to find the probability that more than 5 out of 8 surgeries are successful. This means we want to find P(X>5) for n=8 and p=0.8.
P(X>5) = P(X=6) + P(X=7) + P(X=8)
We can calculate each of these probabilities using the binomial probability formula:
P(X=6) = C(8, 6) * (0.8^6) * ((1-0.8)^(8-6)) = 28 * 0.2621 * 0.04 = 0.2949 P(X=7) = C(8, 7) * (0.8^7) * ((1-0.8)^(8-7)) = 8 * 0.2097 * 0.2 = 0.3355 P(X=8) = C(8, 8) * (0.8^8) * ((1-0.8)^(8-8)) = 1 * 0.1678 * 1 = 0.1678
Adding these probabilities together gives:
P(X>5) = P(X=6) + P(X=7) + P(X=8) = 0.2949 + 0.3355 + 0.1678 = 0.7982
So, the probability that more than 5 out of 8 surgeries are successful is approximately 0.80, or 80%, when rounded to two decimal places.
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