Probability Table Quiz: Probability Table Purpose and Interpretation

P(X < 3) includes all outcomes strictly less than 3, which are x = 1 and x = 2. P(X < 3) = P(X=1) + P(X=2) = 0.3 + 0.5 = 0.8. Alternatively, using the complement: P(X < 3) = 1 minus P(X=3) = 1 minus 0.2 = 0.8. Both methods confirm the answer.

The addition rule states P(A or B) = P(A) + P(B) minus P(A and B). Substituting: P(A1 or B1) = 0.40 + 0.30 minus 0.12 = 0.58. Subtracting P(A1 and B1) is necessary to avoid double-counting outcomes where both A1 and B1 occur together. Without this correction the sum 0.40 + 0.30 = 0.70 would count the intersection twice.

Option A: 0.2 + 0.3 + 0.5 = 1.0 and all values between 0 and 1 — valid. Option B: 0.1 + 0.2 + 0.3 + 0.4 = 1.0 and all values between 0 and 1 — valid. Option C: 0.3 + 0.4 + 0.4 = 1.1, which exceeds 1 — invalid. Option D: 0.6 + 0.4 = 1.0 and both values between 0 and 1 — valid. Only option C fails the requirement that all probabilities must sum to exactly 1.

Two events are independent when P(A and B) = P(A) multiplied by P(B). Checking: P(A1) multiplied by P(B1) = 0.40 multiplied by 0.30 = 0.12. Since P(A1 and B1) = 0.12 also equals 0.12, the independence condition is satisfied. Knowing that B1 occurred provides no information about whether A1 occurred, and vice versa.

The answer is False. The column margins in a joint probability table are the marginal probabilities P(B) for each column event, and they must sum to 1 — not to the number of columns. The column events form a complete partition of the sample space, so their probabilities cover all possible outcomes and must total 1. Similarly the row margins sum to 1, not the number of rows.

A probability of 0 assigned to an outcome means that outcome has zero chance of occurring. It is a valid entry in a probability table — it simply indicates x = 0 is an impossible outcome for this distribution. The remaining probabilities for all other outcomes must still sum to 1. This is distinct from an outcome being absent from the table entirely, though both indicate x = 0 cannot occur.

First compute E[X] = 0(0.1) + 1(0.2) + 2(0.4) + 3(0.3) = 1.9. Then compute E[X squared] = 0 squared (0.1) + 1 squared (0.2) + 2 squared (0.4) + 3 squared (0.3) = 0 + 0.2 + 1.6 + 2.7 = 4.5. Variance = E[X squared] minus (E[X]) squared = 4.5 minus 3.61 = 0.89.

Option A is true: each interior cell gives P(A and B) — the probability that both the row event and column event occur together. Option B is true: row margins are obtained by summing across all columns in a row, giving P(A) for that row event. Option D is true: column margins are obtained by summing down all rows in a column, giving P(B) for that column event. Option C is false: the sum of all interior cells equals 1 because they represent a complete partition of the sample space, not 2.

E[X] = sum of each outcome multiplied by its probability. E[X] = 1(0.3) + 2(0.5) + 3(0.2) = 0.3 + 1.0 + 0.6 = 1.9. Each term represents the contribution of that outcome to the overall average. Since x = 2 has the highest probability of 0.5, the expected value of 1.9 is pulled close to 2.

The answer is True. The requirement that probabilities sum to 1 reflects the certainty that some outcome must occur. A table where probabilities sum to less than 1 is incomplete, and a table where they sum to more than 1 is contradictory. Both conditions — each probability between 0 and 1, and all probabilities summing to 1 — must hold simultaneously for a table to be valid.

The column labeled P(X = x) gives the probability that the random variable X takes the specific value shown in the outcome column of the same row. Each entry is a number between 0 and 1 representing the likelihood of that exact outcome. Reading any row gives you the outcome and its associated probability directly.

The column margin P(B2) is found by summing all interior cells in the B2 column. P(B2) = P(A1 and B2) + P(A2 and B2) = 0.28 + 0.42 = 0.70. Column margins represent the marginal probability of the column event regardless of which row event occurred. Confirming: P(B1) + P(B2) = 0.30 + 0.70 = 1.00.

Conditional probability is defined as P(A1 given B1) = P(A1 and B1) divided by P(B1). Substituting: 0.12 divided by 0.30 equals 0.40. The interior cell 0.12 gives the joint probability and the column margin 0.30 gives the probability of the conditioning event B1. The result 0.40 means that among outcomes where B1 occurs, 40 percent also involve A1.

The answer is False. Individual probabilities can be anywhere between 0 and 1 inclusive, so values greater than 0.5 are perfectly allowed as long as all probabilities together sum to 1. For example a table with P(X=0) = 0.9 and P(X=1) = 0.1 is completely valid. The only requirements are that each individual probability lies between 0 and 1, and all probabilities sum to exactly 1.

Since the outcomes 0 and 3 are mutually exclusive, their probabilities are simply added. P(X=0 or X=3) = P(X=0) + P(X=3) = 0.1 + 0.3 = 0.4. Mutual exclusivity means both outcomes cannot occur simultaneously for a single observation of X, so there is no overlap to subtract.

Option A is valid: each value is between 0 and 1, and 0.2 + 0.3 + 0.5 = 1. Option C is valid: each value is 0.25, all between 0 and 1, and 0.25 multiplied by 4 = 1. Option B is invalid: 0.1 + 0.4 + 0.6 = 1.1, which exceeds 1. Option D is invalid: the value negative 0.1 is less than 0, which violates the requirement that all probabilities must be non-negative.

E[X] = sum of each outcome multiplied by its probability. E[X] = 0(0.1) + 1(0.2) + 2(0.4) + 3(0.3) = 0 + 0.2 + 0.8 + 0.9 = 1.9. The expected value is a weighted average of the outcomes, where each weight is the corresponding probability. It represents the long-run average value of X over many repetitions.

For a probability table to be valid, all probabilities must sum to exactly 1. The known probabilities sum to 0.1 + 0.3 + 0.4 = 0.8. The missing probability equals 1 minus 0.8 equals 0.2. Confirming: 0.1 + 0.3 + 0.4 + 0.2 = 1.0, and all four values are between 0 and 1.

The answer is True. Two events are defined as independent when the occurrence of one does not affect the probability of the other. The mathematical condition for independence is P(A and B) = P(A) multiplied by P(B). In a joint probability table, independence can be verified by checking whether each interior cell equals the product of its corresponding row and column margins.

P(X ≤ 2) is found by summing the probabilities for all outcomes up to and including 2. P(X=0) + P(X=1) + P(X=2) = 0.1 + 0.2 + 0.4 = 0.7. Alternatively, using the complement: P(X ≤ 2) = 1 minus P(X=3) = 1 minus 0.3 = 0.7. Both approaches confirm the answer.