Question 965490: *****Of all the trees planted by a landscaping firm, 55% survive. What is the probability that 12 or more of the 14 trees they just planted will survive?
(Use a table of binomial probabilities. Give your answer correct to four decimal places.)
I have been stuck on this for days!! Please help me. If possible please show step-by-step instructions.
Thanks
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 55% survive, so p = .55
this means that 45% don't survive.
q = 1 - p = 1 - .55 = .45
you have p = .55
you have q = .45
the binomial probability formula is:
p(x) = c(n,x) * p^x * q^(n-x)
x represents the ratio of plants that survive.
n is the total number of trees that were planted.
p(x) is the probability that x will survive.
c(n,x) is the number of possible combinations of n plants taken x at a time where order is not important.
c(n,x) is defined as n! / (x! * (n-x)!)
you want to know the probability that 12 or more will survice.
this is equal to the probability that exactly 12 will survive plus the probability that exactly 13 will survice plus the probability that exactly 14 will survive.
so you need to find p(12) + p(13) + p(14).
p(12) = c(14,12) * p^12 * q^2 = 91 * .55^12 * .45^2 = .0141194797
p(13) = c(14,13) * p^13 * q^1 = 14 * .55^13 * .45^1 = .0026549449
p(14) = c(14,14) * p^14 * q^0 = 1 * .55^14 * .45^0 = .0002317809043
p(12) + p(13) + p(14) = .0170062055
that's your solution.
here's one of the combination calculations to show you how they're done.
c(14,12) = 14! / (12! * 2!) = (14 * 13 * 12!) / (12! * 2!) = (14 * 13) / 2! = 7 * 13 = 91.
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