Probability and Statistics practice questions

Answer: A) 36

• ACorrect. \(\text{Var}(aX) = a^2\text{Var}(X)\).
• BIncorrect. Square the constant.
• CIncorrect. Scaling changes the variance.
• DIncorrect. Multiply by 4.

Answer: B) 0.4

• AIncorrect. Divide, don't multiply.
• BCorrect. \(P(A \cap B)/P(B)\).
• CIncorrect. Divide, don't add.
• DIncorrect. That's inverted.

Answer: C) 0.95

• AIncorrect. That's about |Z| < 2.58.
• BIncorrect. That's |Z| < 1.
• CCorrect. A two-sided 95% interval.
• DIncorrect. That's about |Z| < 1.645.

Answer: C) 2

• AIncorrect. That's the MTBF in hours.
• BIncorrect. That's the hourly rate.
• CCorrect. Expected failures equal λ times t.
• DIncorrect. That's off by a factor of 10.

Answer: A) 0.271

• ACorrect. \(1 - 0.9^3\).
• BIncorrect. Use the complement rule.
• CIncorrect. That's none failing.
• DIncorrect. That's all three failing.

Answer: D) 10.0 V

• AIncorrect. Divide the sum by 3.
• BIncorrect. Recheck the sum.
• CIncorrect. That's the sample standard deviation.
• DCorrect. The sum 30.0 divided by 3.

Answer: A) 2 hours

• ACorrect. Mean = 1/λ.
• BIncorrect. That's the rate.
• CIncorrect. Take 1/λ.
• DIncorrect. Recheck: 1/0.5.

Answer: A) 0.135

• ACorrect. \(P(0) = e^{-2}\).
• BIncorrect. That isn't the Poisson result.
• CIncorrect. That's \(P(1) = 2e^{-2}\).
• DIncorrect. Zero failures is still possible.

Answer: B) 0.75

For the union of two events, use the addition rule: P(A or B) = P(A) + P(B) − P(A and B) = 0.6 + 0.3 − 0.15 = 0.75.

• AIncorrect. This adds P(A) and P(B) but doesn't subtract the overlap, so the outcomes in both events are counted twice.
• BCorrect. By the addition rule, P(A or B) = P(A) + P(B) − P(A and B) = 0.6 + 0.3 − 0.15 = 0.75.
• CIncorrect. Subtracting the joint probability twice (0.9 − 0.30) gives the probability that exactly one event occurs (exclusive or), not the inclusive union.
• DIncorrect. This is P(A)·P(B), which would be the joint probability if the events were independent. It is not the union, and here the given joint probability is 0.15 anyway.

Answer: B) 0.243

Use the binomial formula: P(exactly 1) = C(3,1)p(1 − p)² = 3(0.1)(0.81) = 0.243.

• AIncorrect. This is the probability for a single part. It ignores the requirement that the other two be good and the three ways to choose which part is defective.
• BCorrect. By the binomial distribution, P(X = 1) = C(3,1)(0.1)(0.9)² = 3 × 0.1 × 0.81 = 0.243.
• CIncorrect. This is the probability that at least one part is defective, 1 − 0.9³ = 0.271. It wrongly includes the cases with two or three defective parts.
• DIncorrect. This adds the individual probabilities (3 × 0.1) and leaves out the factor 0.9² for the other two parts being good.