Discrete Distribution Quiz: Discrete Distribution Tables

1) In a discrete distribution table, what does the entry P(X = x) represent for a particular row x?

The probability that the random variable equals that row’s outcome x

The count of times x occurred in a sample

The cumulative probability up to x

The z-score of x

A probability table pairs each outcome x with its probability P(X = x). The value in the P(X = x) column is the chance of that exact outcome on a single trial.

Explanation

A probability table pairs each outcome x with its probability P(X = x). The value in the P(X = x) column is the chance of that exact outcome on a single trial.

2) Which of the following lists could be valid probability columns for a discrete distribution table? Select all that apply.

{0.2, 0.3, 0.5}

{0.1, 0.4, 0.6}

{0, 1, 0}

{0.25, 0.25, 0.25, 0.25}

{−0.1, 0.5, 0.6}

To be valid, probabilities must be between 0 and 1 inclusive and sum to 1. A: 0.2+0.3+0.5=1 (valid). B: 0.1+0.4+0.6=1.1>1 (invalid). C: 0+1+0=1 (valid). D: 0.25×4=1 (valid). E includes −0.1 (invalid).

Explanation

To be valid, probabilities must be between 0 and 1 inclusive and sum to 1. A: 0.2+0.3+0.5=1 (valid). B: 0.1+0.4+0.6=1.1>1 (invalid). C: 0+1+0=1 (valid). D: 0.25×4=1 (valid). E includes −0.1 (invalid).

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3) Given the table x={0,1,2} with probabilities {0.2, 0.5, 0.3}, what is P(X ≥ 2)?

0.2

0.3

0.5

0.8

P(X ≥ 2) includes outcomes {2}. From the table P(2)=0.3, so P(X ≥ 2)=0.3. (If there were outcomes 3,4,…, we would sum those rows too.)

Explanation

P(X ≥ 2) includes outcomes {2}. From the table P(2)=0.3, so P(X ≥ 2)=0.3. (If there were outcomes 3,4,…, we would sum those rows too.)

4) Any probability in a discrete distribution table must be between 0 and 1 inclusive.

True

False

By the axioms of probability, 0 ≤ P(X = x) ≤ 1 for every outcome x. Values outside this range are not permissible.

Explanation

By the axioms of probability, 0 ≤ P(X = x) ≤ 1 for every outcome x. Values outside this range are not permissible.

5) A table lists P(X=0)=0.10, P(X=1)=0.20, P(X=2)=0.40, and P(X=3)=____ so that the distribution is valid.

The probabilities must sum to 1. Sum so far = 0.10+0.20+0.40=0.70, so P(X=3)=1−0.70=0.30.

Explanation

The probabilities must sum to 1. Sum so far = 0.10+0.20+0.40=0.70, so P(X=3)=1−0.70=0.30.

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6) A column of probabilities reads {0.15, 0.25, 0.20, 0.35}. Is this a valid probability distribution?

No, the sum is 0.95 which is not 1

Yes, all entries are ≤ 1

Yes, there are four outcomes

No, because probabilities cannot be decimals

Check the sum: 0.15+0.25+0.20+0.35=0.95. A valid distribution must sum to 1 exactly, so this table is invalid.

Explanation

Check the sum: 0.15+0.25+0.20+0.35=0.95. A valid distribution must sum to 1 exactly, so this table is invalid.

7) A table has outcomes x={1,2,3,4} with P={0.1, 0.2, 0.4, 0.3}. What is P(X in {1,3})?

0.1

0.3

0.4

0.5

Add the probabilities for the requested set: P(1)+P(3)=0.1+0.4=0.5.

Explanation

Add the probabilities for the requested set: P(1)+P(3)=0.1+0.4=0.5.

8) From the table x={1,2,3,4,5} with P={0.10, 0.15, 0.25, 0.30, 0.20}, select all outcomes with probability at least 0.25.

X=1

X=2

X=3

X=4

X=5

Check each probability: P(1)=0.10, P(2)=0.15, P(3)=0.25, P(4)=0.30, P(5)=0.20. Those ≥0.25 are x=3 and x=4.

Explanation

Check each probability: P(1)=0.10, P(2)=0.15, P(3)=0.25, P(4)=0.30, P(5)=0.20. Those ≥0.25 are x=3 and x=4.

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9) In a discrete distribution table, the probabilities for all listed outcomes must add to 1, even if some outcomes have probability 0.

True

False

Completeness requires that the probabilities across all possible outcomes sum to 1. Outcomes with probability 0 still count toward the total set of outcomes.

Explanation

Completeness requires that the probabilities across all possible outcomes sum to 1. Outcomes with probability 0 still count toward the total set of outcomes.

10) Compute the total probability for P-column {0.15, 0.35, 0.25, 0.25}. The sum is __________.

Add: 0.15+0.35=0.50; 0.50+0.25=0.75; 0.75+0.25=1.00. The column sums to 1, so it is valid on this criterion.

Explanation

Add: 0.15+0.35=0.50; 0.50+0.25=0.75; 0.75+0.25=1.00. The column sums to 1, so it is valid on this criterion.

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11) Which set below describes a valid probability column?

{0.3, 0.3, 0.3, 0.1}

{0.2, 0.2, 0.2, 0.2}

{0.1, 0.4, 0.4, 0.2}

{0.5, 0.6, −0.1}

Check sums and signs: A sums to 1 and entries are between 0 and 1 (valid). B sums to 0.8 (invalid). C sums to 1.1 (invalid). D has 0.6>1 and −0.1

Explanation

Check sums and signs: A sums to 1 and entries are between 0 and 1 (valid). B sums to 0.8 (invalid). C sums to 1.1 (invalid). D has 0.6>1 and −0.1

12) A fair discrete distribution over five outcomes assigns the same P(X = x) to each row. What is each probability?

0.1

0.2

0.25

0.5

For 5 equally likely outcomes, total probability 1 is split evenly: 1/5=0.20 for each row.

Explanation

For 5 equally likely outcomes, total probability 1 is split evenly: 1/5=0.20 for each row.

13) Given x={0,1,2,3} with P={0.25, 0.35, 0.15, 0.25}, select all true readings from the table.

P(X=0)=0.25

P(X=1)=0.30

P(X=2)=0.15

P(X in {1,3})=0.60

P(X≥2)=0.40

Read each entry and sum as needed: P(0)=0.25 (true). P(1)=0.35, not 0.30 (false). P(2)=0.15 (true). P({1,3})=0.35+0.25=0.60 (true). P(X≥2)=0.15+0.25=0.40 (true).

Explanation

Read each entry and sum as needed: P(0)=0.25 (true). P(1)=0.35, not 0.30 (false). P(2)=0.15 (true). P({1,3})=0.35+0.25=0.60 (true). P(X≥2)=0.15+0.25=0.40 (true).

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14) If the probabilities in a table sum to 1.05, the table is invalid.

True

False

The total probability across all possible outcomes must be exactly 1. A sum of 1.05 exceeds 1, so the table is invalid.

Explanation

The total probability across all possible outcomes must be exactly 1. A sum of 1.05 exceeds 1, so the table is invalid.

15) A biased coin has P(Tails)=0.37. Fill P(Heads)=________ to complete a valid distribution over {Heads,Tails}.

For two exhaustive outcomes, P(Heads)+P(Tails)=1. Compute P(Heads)=1−0.37=0.63.

Explanation

For two exhaustive outcomes, P(Heads)+P(Tails)=1. Compute P(Heads)=1−0.37=0.63.

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16) A table lists P={0.40, 0.35, 0.25} for outcomes x={0,1,2}. Is this a valid distribution?

Yes, the sum is 1.00 and all entries are in [0,1]

No, there are only three outcomes

No, because 0.40 is too large

No, because probabilities must be equal

Check the sum: 0.40+0.35+0.25=1.00. All entries are within [0,1], so the table is valid.

Explanation

Check the sum: 0.40+0.35+0.25=1.00. All entries are within [0,1], so the table is valid.

17) Using x={0,1,2,3} with P={0.1, 0.3, 0.4, 0.2}, what is P(X ≤ 1)?

0.1

0.3

0.4

0.5

P(X ≤ 1)=P(0)+P(1)=0.1+0.3=0.4. Read the rows for outcomes 0 and 1 and add their probabilities.

Explanation

P(X ≤ 1)=P(0)+P(1)=0.1+0.3=0.4. Read the rows for outcomes 0 and 1 and add their probabilities.

18) Which of the following sets cannot be a probability column? Select all that apply.

{0.6, 0.6}

{0.7, 0.2, 0.1}

{0.3, −0.1, 0.8}

{0.1, 0.1, 0.1, 0.1, 0.7}

{0.05, 0.15, 0.25, 0.20, 0.45}

Check sums and signs: A sum=1.2>1 (invalid). B sum=1 (valid). C contains −0.1 (invalid). D sum=1.1>1 (invalid). E sum=0.05+0.15+0.25+0.20+0.45=1.10>1 (invalid).

Explanation

Check sums and signs: A sum=1.2>1 (invalid). B sum=1 (valid). C contains −0.1 (invalid). D sum=1.1>1 (invalid). E sum=0.05+0.15+0.25+0.20+0.45=1.10>1 (invalid).

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19) For any event A formed by listed outcomes, P(A) equals the sum of P(X = x) over those outcomes in the table.

True

False

By additivity for mutually exclusive outcomes, the probability of a set of outcomes is the sum of the probabilities of its disjoint elements.

Explanation

By additivity for mutually exclusive outcomes, the probability of a set of outcomes is the sum of the probabilities of its disjoint elements.

20) In the table x={3,4,5,6} with probabilities P={0.10, 0.25, ____, 0.30}, fill P(X=5) so the distribution is valid.

Sum so far = 0.10 + 0.25 + 0.30 = 0.65. The missing value must be 1 − 0.65 = 0.35 so that the probabilities add to 1.

Explanation

Sum so far = 0.10 + 0.25 + 0.30 = 0.65. The missing value must be 1 − 0.65 = 0.35 so that the probabilities add to 1.

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