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Assume that a procedure yields a binomial distribution with a trial that is repeated 5 times. use the binomial probability of 2 successes given that a single success has a probability of 0.712.

a. 0.205
b. 0.121
c. 0.012
d. 0.879

User Alavalathi
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1 Answer

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Answer: The probability of 2 successes where the single success has a probability of 0.712 and the number of trials 5 will be B 0.121

Explanation:

n = number of trials

x = number of successes

p = probability of getting success in one trial

n = 5

x = 2

p = 0.712

You should be able to plug these values into your graphing calculator to find the answer.

Use BinomPDF as BinomPDF is the probability that there will be X successes in n trials if there is a probability p of success for each trial.

If you have a Ti-84

Push 2nd, then vars. Scroll down until you see BinomPDF and then click on it. From there plug in the values and push paste.

The output should be P ( 2 ) = 0.121098133

Which can be rounded to P ( 2 ) ≈ 0.121

User James Simpson
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