Using a Distribution to Find Probabilities In Exercises 11–26,...

Question

Using a Distribution to Find Probabilities In Exercises 11–26, find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine whether the events are unusual. If convenient, use a table or technology to find the probabilities.

Pass Completions NFL player Aaron Rodgers completes a pass 65.1% of the time. Find the probability that (a) the first pass he completes is the second pass, (b) the first pass he completes is the first or second pass, and (c) he does not complete his first two passes. (Source: National Football League)

Accepted Answer

All right. Hello, everyone. So this question says, a customer service agent successfully resolves a caller's issue on the first attempt 72% of the time. Find the probability that for part one. The 1st successful resolution occurs on the 3rd call. For part two, the first successful resolution occurs on or before the second call. And for part 3, the agent does not successfully resolve any of the 1st 3 calls. So, in this case, we're specifically told that the probability of success, that is resolving an issue on a single call or P is equal to 0.72% or 72%. Now, this question is an example of a geometric distribution, because we're looking for the probability that the first success occurs on a specific attempt. So, the geometric probability formula, if you recall, is P X equals 1 subtracted by P. To the power of X subtracted by 1 multiplied by P. So, for part one, for example, if we want to calculate the probability that the first successful resolution occurs on the 3rd call, X is equal to 3. So P of 3. Is equal to 1 subtracted by 0.72. To the power of 3 subtracted by 1, multiplied by 0.72. And so this equals 0.0564 as the answer to part one. Now for part two In this case, it's saying that the first successful resolution occurs on or before the 2nd call. So, that means it's the first success occurs either on the 1st or 2nd call. So, The probability of X. Less than or equal to 2. Would therefore be equal to the probability of one. Added to the probability of 2. And so, here for the probability of the success being on the first call, that would be 0.28. Or one subtracted by 0.72 in this case of the power of zero. Multiplied by 0.72. And then this would be added to. 0.28 to the first power multiplied by 0.72. So ultimately, the answer to part two would be 0.9216. This leads us now to part 3. Which is the probability that there are no successes in the 1st 3 calls. So, if you want to find the probability that none of the 1st 3 calls resulted in a success, what you could do. I take one subtracted by P. That is one subtracted by 0.72. And then simply cube this probability. Because subtracting 0.72 from 1 would give you the probability of failure per each attempt. So, 0.28 cubed ultimately gives you approximately 0.0220. As the answer to part three. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

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Key Concepts

Geometric Distribution

Binomial Distribution

Unusual Events

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