2026 ELITE CERTIFICATION PROTOCOL

Probability Practice Test 2026 | Exam Prep

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Q1Domain Verified
A biased coin has a probability of landing heads P(H) = 0.7. If the coin is tossed 5 times, what is the probability of getting exactly 3 heads, given that the first toss was a head?
C(4,3) * (0.7)^3 * (0.3)^2
C(4,2) * (0.7)^3 * (0.3)^2
C(4,2) * (0.7)^2 * (0.3)^2
C(5,3) * (0.7)^3 * (0.3)^2
Q2Domain Verified
Consider two events A and B such that P(
= 0.5, and P(A ∪ B) = 0.8. What is P(A' ∩ B')? A) 0.1 B) 0.2
0.3
= 0.6, P(
0.4
Q3Domain Verified
tests the understanding of De Morgan's laws and the relationship between union and intersection probabilities. We know that P(A ∪ B) = P(
+ P(
0.5
0.7
- P(A ∩ B). Substituting the given values, we get 0.8 = 0.6 + 0.5 - P(A ∩ B), which implies P(A ∩ B) = 1.1 - 0.8 = 0.3. By De Morgan's law, P(A' ∩ B') = P((A ∪ B)'). The probability of the complement of an event is 1 minus the probability of the event, so P((A ∪ B)') = 1 - P(A ∪ B). Therefore, P(A' ∩ B') = 1 - 0.8 = 0.2. Option A is incorrect because it might arise from incorrectly calculating P(A ∩ B) or misapplying De Morgan's law. Option C is incorrect as it is the probability of the intersection of A and B, not the intersection of their complements. Option D is incorrect and does not logically follow from the given probabilities or standard probability rules. Question: A continuous random variable X follows a uniform distribution on the interval [0, 10]. What is the probability that X is greater than 3 and less than 7, i.e., P(3 < X < 7)? A) 0.3 B) 0.4

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This domain protocol is rigorously covered in our 2026 Elite Framework. Every mock reflects direct alignment with the official assessment criteria to eliminate performance gaps.

This domain protocol is rigorously covered in our 2026 Elite Framework. Every mock reflects direct alignment with the official assessment criteria to eliminate performance gaps.

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