Question 1065962: What is the answer for question 133479?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe your answer can be found here:
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.133479.html
based on this, the solution would be p(x=3) = .193536 and p(x>4) = p(x=5) + p(x=6) + p(x=7) = .4199...
the formula to use is p(x) = c(n,x) * p^x * q^(n-x)
x is the number of occurrences you are looking for.
p is the probability of success.
q is the probability of failure which is equal to 1 - p(x).
c(n,x) is the number of possible ways you can get x things fr0om n things when order is not important.
you have:
n = 7
p = .6
q = .4
for example:
p(x=3) is equal to c(7,3) * .6^3 * .4^4
c(7,3) = 35
formula becomes p(x=3) = 35 * .6^3 * .4^4
solve for p(x=3) to get p(x=3) = .193536
the formula for c(n,x) is n! / (x! * (n-x)!)
when n = 7 and x = 3, this formula becomes c(7,3) = 7! / (3! * 4!).
this becomes c(7,3) = (7*6*5*4*3*2*1) / (3*2*1*4*3*2*1).
factor out 4*3*2*1 and this becomes c(7,3) = (7*6*5)/(3*2*1)
solve to get c(7,3) = 7*5 = 35
you can do the same for p(5) and p(6) and p(7) and you will get p(x>4) is equal to the sum of those probabilities.
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