Discrete Distribution Quiz: Discrete Distribution Tables

Checking each option: Option A: 0.3+0.3+0.3+0.1 = 1.0 and all entries are between 0 and 1 — valid. Option B: 0.2+0.2+0.2+0.2 = 0.8, which is less than 1 — invalid. Option C: 0.1+0.4+0.4+0.2 = 1.1, which exceeds 1 — invalid. Option D: contains negative 0.1 which is below 0 — invalid. Only option A satisfies both validity requirements.

P(X ≥ 3) includes outcomes x=3 and x=4. P(X=3) + P(X=4) = 0.3 + 0.1 = 0.4. Alternatively using the complement: P(X ≥ 3) = 1 minus P(X ≤ 2) = 1 minus (0.1+0.2+0.3) = 1 minus 0.6 = 0.4. Both methods confirm the answer.

The answer is True. This follows directly from the additivity axiom of probability: the probability of a union of mutually exclusive events equals the sum of their individual probabilities. Since each outcome in a discrete distribution is mutually exclusive with every other outcome, the probability of any collection of outcomes is simply the sum of their individual probabilities. This is the foundation of all probability calculations using a distribution table.

Checking each option: Option A: 0.25+0.25+0.5 = 1.0, all entries in [0,1] — valid. Option B: 0.7+0.2+0.1 = 1.0, all entries in [0,1] — valid. Option C: 0.6+0.6 = 1.2, which exceeds 1 — invalid. Option D: 0.4+0.3+0.3 = 1.0, all entries in [0,1] — valid. Only option C fails because its probabilities sum to more than 1.

E[X] = sum of each outcome multiplied by its probability. E[X] = 1(0.1) + 2(0.2) + 3(0.4) + 4(0.3) = 0.1 + 0.4 + 1.2 + 1.2 = 2.9. Since the highest probabilities belong to x=3 and x=4, the expected value of 2.9 is pulled toward the upper end of the distribution.

Checking the sum: 0.40 + 0.35 + 0.25 = 1.00. Every entry is between 0 and 1 inclusive. Both validity conditions are satisfied, so the table is valid. Option B is incorrect — there is no minimum number of outcomes required. Option C is incorrect — individual probabilities can be any value between 0 and 1 as long as all entries sum to 1. Option D is incorrect — outcomes do not need equal probabilities.

The mode of a discrete distribution is the outcome with the highest individual probability. Checking each: P(1)=0.10, P(2)=0.20, P(3)=0.40, P(4)=0.30. The outcome x=3 has the largest probability of 0.40 and is therefore the mode. The mode is the single most likely outcome on any given observation of X.

The answer is True. The total probability across all possible outcomes must equal exactly 1. A sum of 1.05 exceeds 1, which violates one of the fundamental axioms of probability. This would imply the combined chance of all outcomes is greater than certainty, which is impossible. A sum below 1 is equally invalid — the table would be incomplete.

Option A is true: the table directly shows P(X=0) = 0.25. Option B is false: P(X=1) = 0.35, not 0.30 — reading the wrong value is a common error. Option C is true: P(X=2) = 0.15 as listed. Option D is true: P(X∈{1,3}) = P(1)+P(3) = 0.35+0.25 = 0.60.

For a fair (uniform) distribution, total probability 1 is divided equally among all outcomes. With 5 outcomes, each probability equals 1 divided by 5 equals 0.20. Option A (0.1) would give a total of 0.5. Option C (0.25) would give a total of 1.25. Option D (0.5) would give a total of 2.5. Only option B produces the correct total of 5 multiplied by 0.20 equals 1.

E[X] = sum of each outcome multiplied by its probability. E[X] = 2(0.1) + 4(0.3) + 6(0.4) + 8(0.2) = 0.2 + 1.2 + 2.4 + 1.6 = 5.4. The expected value is the long-run average of X over many repetitions. It does not have to be one of the listed outcomes — here 5.4 falls between the outcomes 4 and 6.

The event 1 < X ≤ 3 includes outcomes strictly greater than 1 and at most 3, which means x = 2 and x = 3. P(X=2) + P(X=3) = 0.15 + 0.25 = 0.40. Note that x = 1 is excluded because the inequality is strict on the left — greater than 1, not greater than or equal to 1.

The answer is True. Assigning a probability of 0 to an outcome is permitted — it simply means that outcome is impossible for this distribution. The table remains valid as long as all listed probabilities are between 0 and 1 inclusive and the total equals 1. Including impossible outcomes explicitly can be useful for completeness, particularly when comparing distributions that differ in which outcomes are possible.

Checking each probability: P(1) = 0.10, which is less than 0.25. P(2) = 0.15, which is less than 0.25. P(3) = 0.25, which equals 0.25 and meets the threshold. P(4) = 0.30, which exceeds 0.25 and meets the threshold. P(5) = 0.20, which is less than 0.25. Only outcomes x = 3 and x = 4 have probabilities at least 0.25.

To find the probability that X belongs to the set {1, 3}, add the individual probabilities for those outcomes. P(X=1) + P(X=3) = 0.1 + 0.4 = 0.5. The outcomes 1 and 3 are mutually exclusive — X cannot equal both simultaneously — so their probabilities are simply summed with no adjustment needed.

Checking the sum: 0.15 + 0.25 + 0.20 + 0.35 = 0.95. A valid distribution requires all probabilities to sum to exactly 1. Since 0.95 does not equal 1, the table is invalid regardless of whether individual entries are between 0 and 1. Option B is a common error — satisfying 0 ≤ P ≤ 1 for each entry is necessary but not sufficient; the sum must also equal 1.

Using the complement rule: P(X ≠ 2) = 1 minus P(X=2) = 1 minus 0.15 = 0.85. The complement of the event X=2 is the event that X takes any value other than 2, which includes outcomes 0, 1, and 3. Confirming directly: P(0)+P(1)+P(3) = 0.25+0.35+0.25 = 0.85.

The answer is False. The mode of a discrete distribution is the outcome with the highest individual probability P(X=x), not the highest cumulative probability. Cumulative probability adds up probabilities from the lowest outcome to a given point and always reaches 1 at the highest outcome. For example in the distribution x={1,2,3} with P={0.2,0.5,0.3}, the mode is x=2 because P(2)=0.5 is the largest individual probability.

The even outcomes in this distribution are x = 2 and x = 4. P(X is even) = P(X=2) + P(X=4) = 0.2 + 0.3 = 0.5. The probability of a set of mutually exclusive outcomes is found by summing their individual probabilities. The odd outcomes x = 1 and x = 3 account for the remaining probability of 0.1 + 0.4 = 0.5, confirming the two groups partition the total probability of 1.

Checking each option: Option A: E[X] = 1(0.3)+2(0.3)+3(0.2)+4(0.2) = 0.3+0.6+0.6+0.8 = 2.3. Option B: E[X] = 0(0.1)+1(0.2)+2(0.5)+3(0.2) = 0+0.2+1.0+0.6 = 1.8. Option C: E[X] = 0(0.25)+2(0.5)+4(0.25) = 0+1.0+1.0 = 2.0. Option D: E[X] = 1(0.2)+2(0.5)+3(0.3) = 0.2+1.0+0.9 = 2.1. Only option C gives exactly E[X] = 2.