Probability Table Quiz: Probability Table Purpose And Interpretation

1) What does a probability distribution table primarily show for a discrete random variable X?

Each possible outcome x and its probability P(X = x)

Only the most likely outcome

Only the mean and median of X

A list of sample data without probabilities

A probability table pairs each outcome x with P(X = x). This lets you read the chance of any exact outcome directly from its row.

Explanation

A probability table pairs each outcome x with P(X = x). This lets you read the chance of any exact outcome directly from its row.

2) Select all statements that must be true for any probability distribution table.

Every probability is between 0 and 1 inclusive

All listed probabilities add up to 1

Outcomes must be mutually exclusive (no overlap)

There must be at least 5 outcomes

Probabilities may be negative if rare

Valid tables require 0 ≤ P ≤ 1 (A), total probability across distinct outcomes equals 1 (B), and outcomes are mutually exclusive so probabilities don’t double-count (C).

Explanation

Valid tables require 0 ≤ P ≤ 1 (A), total probability across distinct outcomes equals 1 (B), and outcomes are mutually exclusive so probabilities don’t double-count (C).

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3) Consider the table: x = {0,1,2} with P = {0.2, 0.5, 0.3}. What does the row for x = 1 tell you?

P(X = 1) = 0.5

P(X = 1) = 0.2

P(X = 1) = 0.3

P(X = 1) = 0.7

Match the outcome x = 1 to its probability column entry 0.5. So the chance of X being 1 is 0.5.

Explanation

Match the outcome x = 1 to its probability column entry 0.5. So the chance of X being 1 is 0.5.

4) If a probability table lists probabilities 0.1, 0.2, 0.3, and 0.4 for all outcomes, it is valid.

True

False

Add the probabilities: 0.1 + 0.2 + 0.3 + 0.4 = 1. Since each is between 0 and 1 and the sum is 1, the table is valid.

Explanation

Add the probabilities: 0.1 + 0.2 + 0.3 + 0.4 = 1. Since each is between 0 and 1 and the sum is 1, the table is valid.

5) A table lists outcomes {A,B,C} with P(A)=0.25, P(B)=0.5, and P(C)=______. Fill the blank so the table is valid.

To be valid, probabilities must sum to 1. Compute P(C) = 1 − [0.25 + 0.5] = 1 − 0.75 = 0.25.

Explanation

To be valid, probabilities must sum to 1. Compute P(C) = 1 − [0.25 + 0.5] = 1 − 0.75 = 0.25.

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6) In a two‑column table with column 1 labeled 'Outcome x' and column 2 labeled 'P(X = x)', what does column 2 represent?

The probability of each specific outcome x

A count of how often x occurred

The cumulative frequency of x

The standard score (z) of each x

Column 2 lists P(X = x) for each outcome in column 1. Each entry is the probability of that exact outcome.

Explanation

Column 2 lists P(X = x) for each outcome in column 1. Each entry is the probability of that exact outcome.

7) Given x={1,2,3,4} and P={0.1, 0.3, 0.4, 0.2}, what is P(X ≥ 3)?

0.3

0.4

0.6

0.7

P(X ≥ 3) = P(3) + P(4) = 0.4 + 0.2 = 0.6. Sum the rows corresponding to 3 and 4.

Explanation

P(X ≥ 3) = P(3) + P(4) = 0.4 + 0.2 = 0.6. Sum the rows corresponding to 3 and 4.

8) In a joint probability table for row events A and column events B, what does an interior cell represent?

P(A and B)

P(A) only

P(B) only

P(A or B)

An interior cell corresponds to P(A ∩ B): the probability that the row event A and the column event B occur together.

Explanation

An interior cell corresponds to P(A ∩ B): the probability that the row event A and the column event B occur together.

9) In a joint probability table, what do the row totals (margins) represent?

Marginal probabilities P(A) for the row events

Conditional probabilities P(A|B)

Joint probabilities P(A and B)

The mean of A

Row margins are obtained by summing across columns and give P(A). Column margins give P(B).

Explanation

Row margins are obtained by summing across columns and give P(A). Column margins give P(B).

10) If a joint table gives P(A and B) and the column margin gives P(B) > 0, how do you compute P(A | B)?

P(A | B) = P(A and B) / P(B)

P(A | B) = P(A) + P(B)

P(A | B) = P(B) / P(A and B)

P(A | B) = P(A) × P(B)

Conditional probability uses P(A | B) = P(A ∩ B) / P(B), where the numerator is the interior cell and the denominator is the B column margin.

Explanation

Conditional probability uses P(A | B) = P(A ∩ B) / P(B), where the numerator is the interior cell and the denominator is the B column margin.

11) A probability table that includes any probability greater than 1 is invalid, even if the total sum is 1.

True

False

Every listed probability must satisfy 0 ≤ P ≤ 1. Any entry above 1 violates probability axioms regardless of the total.

Explanation

Every listed probability must satisfy 0 ≤ P ≤ 1. Any entry above 1 violates probability axioms regardless of the total.

12) A table shows x={0,1,2,3} with P={0.2,0.3,0.25,0.25}. What is P(X ≥ 1)?

0.55

0.6

0.75

0.8

P(X ≥ 1) = 1 − P(0) = 1 − 0.2 = 0.8. Equivalently, sum P(1)+P(2)+P(3)=0.3+0.25+0.25=0.8.

Explanation

P(X ≥ 1) = 1 − P(0) = 1 − 0.2 = 0.8. Equivalently, sum P(1)+P(2)+P(3)=0.3+0.25+0.25=0.8.

13) How does a probability distribution table differ from a frequency table of raw data?

It shows probabilities P(X = x) that sum to 1

It lists only observed counts, not probabilities

It must have exactly 100 outcomes

It cannot be used to compute expectations

A probability table documents P(X = x) values that sum to 1, while a frequency table lists counts. Probabilities can come from models or normalized counts.

Explanation

A probability table documents P(X = x) values that sum to 1, while a frequency table lists counts. Probabilities can come from models or normalized counts.

14) In a joint table, the row for A1 has interior cell probabilities 0.1, 0.2, and 0.15 across three columns. The row total P(A1) is ______.

Row total is the marginal probability for A1: 0.1 + 0.2 + 0.15 = 0.45.

Explanation

Row total is the marginal probability for A1: 0.1 + 0.2 + 0.15 = 0.45.

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15) A distribution table lists P={0.25, 0.25, 0.25, 0.15}. Is it valid and why?

No, because the probabilities sum to 0.9, not 1

Yes, because each entry is ≤ 1

Yes, because there are four outcomes

No, because probabilities cannot be decimals

Sum = 0.25 + 0.25 + 0.25 + 0.15 = 0.90. For validity, total probability across all outcomes must equal 1.

Explanation

Sum = 0.25 + 0.25 + 0.25 + 0.15 = 0.90. For validity, total probability across all outcomes must equal 1.

16) Which labels typically appear in a simple discrete distribution table and how should they be interpreted? Select all that apply.

Outcome (value) column lists each x

Probability column lists P(X = x) for each row

A cumulative column is required for validity

A total row checks sum = 1

A notes column changes probabilities

The outcome column pairs with a probability column (A,B). A final total row (or check) verifies the sum is 1 (D). Cumulative columns and notes are optional and do not set probabilities.

Explanation

The outcome column pairs with a probability column (A,B). A final total row (or check) verifies the sum is 1 (D). Cumulative columns and notes are optional and do not set probabilities.

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17) Given x={0,1,2} and P={0.2,0.5,0.3}, what is the expected value E[X]?

1

1.1

1.2

1.3

E[X] = Σ x·P(x) = 0·0.2 + 1·0.5 + 2·0.3 = 0 + 0.5 + 0.6 = 1.1. Multiply each row’s outcome by its probability and sum.

Explanation

E[X] = Σ x·P(x) = 0·0.2 + 1·0.5 + 2·0.3 = 0 + 0.5 + 0.6 = 1.1. Multiply each row’s outcome by its probability and sum.

18) In a joint probability table, an interior cell at row A2 and column B3 gives P(A2 and B3).

True

False

Joint tables’ interior cells represent intersections of row and column events: P(A2 ∩ B3).

Explanation

Joint tables’ interior cells represent intersections of row and column events: P(A2 ∩ B3).

19) A table lists x={1,2,3,4} with P={0.1,0.2,0.5,0.2}. What is P(X in {2,4})?

0.1

0.2

0.3

0.4

Sum the probabilities for the requested set: P(2)+P(4) = 0.2 + 0.2 = 0.4.

Explanation

Sum the probabilities for the requested set: P(2)+P(4) = 0.2 + 0.2 = 0.4.

20) A distribution lists P(x=0)=0.15, P(x=1)=0.35, P(x=2)=0.25, and P(x=3)=____. Fill the blank so the table is valid.

Sum so far = 0.15 + 0.35 + 0.25 = 0.75. The remaining probability is 1 − 0.75 = 0.25 so the total equals 1.

Explanation

Sum so far = 0.15 + 0.35 + 0.25 = 0.75. The remaining probability is 1 − 0.75 = 0.25 so the total equals 1.

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Probability Table Quiz: Probability Table Purpose and Interpretation