SOLUTION: 8% of all Americans live in poverty. If 37 Americans are randomly selected, find the probability that
a. Exactly 3 of them live in poverty________
b. At most 5 of them live
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Question 1199818: 8% of all Americans live in poverty. If 37 Americans are randomly selected, find the probability that
a. Exactly 3 of them live in poverty________
b. At most 5 of them live in poverty_________
c. At least 4 of them live in poverty_________
d. Between 2 and 6 (including 2 and 6) of them live in poverty_______
Answer by Solver92311(821) (Show Source): You can put this solution on YOUR website!
The probability of exactly 
successes in 
independent trials with a uniform probability of success on any given trial of 
is given by:
\ =\ {{n}\choose{k}}\,(p)^k(1\,-\,p)^{n-k})
where 
is the number of combinations of things taken at a time.
The probability of at most successes in independent trials with a uniform probability of success on any given trial of is given by the sum:
\ =\ \sum_{r=0}^k\,{{n}\choose{r}}\,(p)^r(1\,-\,p)^{n-r})
Which is the probability of exactly zero successes plus the probability of exactly one and so on up to the probability of exactly successes.
The probability of at least successes in independent trials with a uniform probability of success on any given trial of is given by the sum:
\ =\ \sum_{r=k}^n\,{{n}\choose{r}}\,(p)^r(1\,-\,p)^{n-r})
Which is the probability of exactly successes plus the probability of exactly 
and so on up to the probability of exactly n successes.
Just plug in your numbers, remembering to express the probability as a decimal, and do the indicated arithmetic.
It may save you some arithmetic if you recognize that \ =\ 1\ -\ P(n,\geq\,k+1,p))
John
My calculator said it, I believe it, that settles itp
From
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