Question 1167587: Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. Find the probability that at least three of the 25 patients actually have the flu.
Thanks!!
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The probability of  successes in  independent trials where the probability of success on any individual trial is  is given by:
\ =\ {{n}\choose{k}}\,\(p\)^k\,\(1\,-\,p\)^{n-k})
To get the probability of at least 3 successes in 25 trials, you would need to add \ +\ P(=4,25,0.04)\ +\ ...\ +\ P(=25,25,0.04)) ,
a sum with 23 terms, which is to say:
\ =\ \sum_{k=3}^{25}\, {{25}\choose{k}}\,\(p\)^k\,\(1\,-\,p\)^{n-k})
This is a great deal of arithmetic indeed. Fortunately, there is a shortcut. Noting that if you added all of the possible outcomes from 0 successes to 25 successes, you would get a sum of 1 (it is certain that you will get at least zero successes), then you could add the probabilities of exactly zero, exactly 1, and exactly 2 successes, subtract that sum from 1 and achieve the probability of at least 3.
\ =\ 1\ - \sum_{k=0}^{2}\, {{25}\choose{k}}\,\(p\)^k\,\(1\,-\,p\)^{n-k})
A sum with only 3 terms that needs to be subtracted from 1.
John
My calculator said it, I believe it, that settles it
I > Ø
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