5/28/25, 10:37 PM Math 110 - Module 4 Exam
Math 110 - Module 4 Exam
Portage Learning
Question 1
A factory has eight safety systems. During an emergency, the probability of any one of the safety
systems failing is 0.08. What is the probability that six or more safety systems will fail during an
emergency?
Solution:
Use the binomial probability formula:
P(X = x) = C(n, x) * (p^x) * (1-p)^(n-x)
Where: n = 8, p = 0.08, x = 6, 7, 8.
For X = 6: P(X = 6) = 6.21 × 10⁻⁶
For X = 7: P(X = 7) = 1.54 × 10⁻⁷
For X = 8: P(X = 8) = 1.68 × 10⁻⁹
Total Probability: P(X ≥ 6) = (6.21 × 10⁻⁶) + (1.54 × 10⁻⁷) + (1.68 × 10⁻⁹)
= 6.36 × 10⁻⁶.
Answer: The probability is 6.36 × 10⁻⁶.
Question 2
Find each of the following probabilities:
a. P(Z ≤ 1.27)
b. P(Z ≥ -0.73)
c. P(-0.09 ≤ Z ≤ 0.86)
Solution:
a. P(Z ≤ 1.27) = 0.89796.
b. P(Z ≥ -0.73) = 0.7673.
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Math 110 - Module 4 Exam
Portage Learning
Question 1
A factory has eight safety systems. During an emergency, the probability of any one of the safety
systems failing is 0.08. What is the probability that six or more safety systems will fail during an
emergency?
Solution:
Use the binomial probability formula:
P(X = x) = C(n, x) * (p^x) * (1-p)^(n-x)
Where: n = 8, p = 0.08, x = 6, 7, 8.
For X = 6: P(X = 6) = 6.21 × 10⁻⁶
For X = 7: P(X = 7) = 1.54 × 10⁻⁷
For X = 8: P(X = 8) = 1.68 × 10⁻⁹
Total Probability: P(X ≥ 6) = (6.21 × 10⁻⁶) + (1.54 × 10⁻⁷) + (1.68 × 10⁻⁹)
= 6.36 × 10⁻⁶.
Answer: The probability is 6.36 × 10⁻⁶.
Question 2
Find each of the following probabilities:
a. P(Z ≤ 1.27)
b. P(Z ≥ -0.73)
c. P(-0.09 ≤ Z ≤ 0.86)
Solution:
a. P(Z ≤ 1.27) = 0.89796.
b. P(Z ≥ -0.73) = 0.7673.
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