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Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to f...Asked by boof
Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. Round to three decimal places.
n =12, x = 5, p = 0.25
(1 point)
Responses
0.103
0.103
0.027
0.027
0.082
0.082
0.091
n =12, x = 5, p = 0.25
(1 point)
Responses
0.103
0.103
0.027
0.027
0.082
0.082
0.091
Answers
There are no human answers yet.
Answered by
GPT 3.5
The binomial probability formula is given by:
P(x) = (nCx) * p^x * (1-p)^(n-x),
where n is the number of trials, x is the number of successes, p is the probability of success on a single trial, and nCx represents the number of combinations of n items taken x at a time.
Using the given values: n = 12, x = 5, p = 0.25, we can substitute them into the formula:
P(5) = (12C5) * 0.25^5 * (1-0.25)^(12-5)
Calculating (12C5):
(12C5) = 12! / (5! * (12-5)!) = 792
Substituting the values:
P(5) = 792 * 0.25^5 * 0.75^7
Calculating:
P(5) ≈ 0.091
Therefore, the probability of 5 successes in 12 trials with a success probability of 0.25 is approximately 0.091.
P(x) = (nCx) * p^x * (1-p)^(n-x),
where n is the number of trials, x is the number of successes, p is the probability of success on a single trial, and nCx represents the number of combinations of n items taken x at a time.
Using the given values: n = 12, x = 5, p = 0.25, we can substitute them into the formula:
P(5) = (12C5) * 0.25^5 * (1-0.25)^(12-5)
Calculating (12C5):
(12C5) = 12! / (5! * (12-5)!) = 792
Substituting the values:
P(5) = 792 * 0.25^5 * 0.75^7
Calculating:
P(5) ≈ 0.091
Therefore, the probability of 5 successes in 12 trials with a success probability of 0.25 is approximately 0.091.
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