MATH 110 Module 4 Exam
Exam Page 1
A factory has eight safety systems. During an emergency, the probability of any one of the safety
systems failing is .08. What is the probability that six or more safety systems will fail during an
emergency?
f(x) = ( (n!) / (x!(n-x)!) ) x ( (p^x) x ((1-p)^n-x)) )
n=8
x = 6, 7, 8 (number of failures)
p = 0.08
6 failures:
n=8
x=6
p = 0.8
n-x = 8-6 = 2
( (8!) / (6!(2)!) ) x ( (0.08^6) x ((1-0.08)^2)) ) = 6.2 x 10^-6
7 failures:
n=8
x=7
p = 0.8
n-x = 8-7 = 1
( (8!) / (7!(1)!) ) x ( (0.08^7) x ((1-0.08)^1)) ) = 1.54 x 10^-7
8 failures:
n=8
x=8
p = 0.8
n-x = 8-8 = 0
( (8!) / (8!(0)!) ) x ( (0.08^8) x ((1-0.08)^0)) ) = 1.68 x 10^-9
f(6) = 6.21 x 10^-6
, f(7) = 1.54 x 10^-7
f(8) = 1.68 x 10^-9
(6.21 x 10^-6) + (1.54 x 10^-7) + (1.68 x 10^-9) = 6.355x10^-6
Probability of 6,7, and 8 failing during an emergency = 6.36 x 10^-6
Answer Key
A factory has eight safety systems. During an emergency, the probability of any one of the safety
systems failing is .08. What is the probability that six or more safety systems will fail during an
emergency?
Exam Page 2
Find each of the following probabilities:
a. Find P(Z ≤ 1.27) .
P(Z ≤ 1.27) = 0.89796
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