Try a binomial probability table or use a binomial probability function, which is this:
P(x) = (nCx)(p^x)[q^(n-x)]
For 1), find P(0):
x = 0
n = 20
p = .09
q = .91 (q is 1-p)
For 2), find P(1)
x = 1
n,p,q same as 1)
For 3), find P(3)
x = 3
n,p,q same as 1)
For 4), take 1 - P(0).
I'll let you take it from here.
Suppose that each student in stat. has a 9% chance of missing class on any given day and that student attendance is independent. In a stat. class of 20 students, find the probability that:
1) no student is absent =15.16
2) one student is absent =30
3) three student is absent =16.72
4) at least one student is absent =31
1 answer