College Statistics Quiz: Basic Probability Rules

20 questions · exam conditions

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Basic Probability RulesQuestion 1 of 20

A card is drawn from a standard 52-card deck. Let H be the event the card is a Heart, and let F be the event the card is a face card (Jack, Queen, or King). What is $P(H \cup F)$ ?

21/52

28/52

25/52

22/52

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College Statistics Quiz

College Statistics Quiz: Basic Probability Rules

Practice Basic Probability Rules in College Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Basic Probability Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A card is drawn from a standard 52-card deck. Let H be the event the card is a Heart, and let F be the event the card is a face card (Jack, Queen, or King). What is $P(H \cup F)$ ?

21/52
28/52
25/52
22/52 (correct answer)

Explanation: When you encounter a union probability question like this, you're dealing with the probability that at least one of two events occurs. The key insight is recognizing that simply adding the individual probabilities will double-count any overlap between the events. To find

P(H \cup F)

, we use the formula:

P(H \cup F) = P(H) + P(F) - P(H \cap F)

. First, let's identify each component. There are 13 hearts in a standard deck, so

P(H) = \frac{13}{52}

. There are 12 face cards total (3 face cards in each of 4 suits), so

P(F) = \frac{12}{52}

. The intersection

P(H \cap F)

represents cards that are both hearts AND face cards—that's the Jack, Queen, and King of hearts, so

P(H \cap F) = \frac{3}{52}

. Therefore:

P(H \cup F) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52}

, which is answer D. Answer A (21/52) likely comes from incorrectly calculating the overlap as 4 instead of 3. Answer B (28/52) results from forgetting to subtract the intersection entirely—just adding 13 + 12 + 3. Answer C (25/52) suggests adding the probabilities without accounting for overlap:

\frac{13}{52} + \frac{12}{52} = \frac{25}{52}

. Remember: whenever you see "or" in probability (union), always check for overlap between events. The inclusion-exclusion principle prevents double-counting and is essential for union problems.

Question 2

A system relies on two independent components, C1 and C2. The probability that C1 works is 0.9, and the probability that C2 works is 0.8. The system works if at least one of the components works. What is the probability that the system fails?

0.02 (correct answer)
0.18
0.28
0.72

Explanation: Let W1 be the event C1 works, and W2 be the event C2 works. We are given

P(W1)=0.9

and

P(W2)=0.8

. The system fails only if BOTH components fail. Let F1 be the event C1 fails and F2 be the event C2 fails. These are the complements of W1 and W2. So,

P(F1) = 1 - P(W1) = 1 - 0.9 = 0.1

, and

P(F2) = 1 - P(W2) = 1 - 0.8 = 0.2

. Since the components are independent, the events F1 and F2 are also independent. The probability that both fail is

P(F1 \cap F2) = P(F1) \cdot P(F2) = 0.1 \cdot 0.2 = 0.02

Question 3

A card is drawn from a standard 52-card deck. Let H be the event the card is a Heart, and let F be the event the card is a face card (Jack, Queen, or King). What is $P(H \cup F)$ ?

21/52
28/52
25/52
22/52 (correct answer)

P(H \cup F)

, we use the formula:

P(H \cup F) = P(H) + P(F) - P(H \cap F)

. First, let's identify each component. There are 13 hearts in a standard deck, so

P(H) = \frac{13}{52}

. There are 12 face cards total (3 face cards in each of 4 suits), so

P(F) = \frac{12}{52}

. The intersection

P(H \cap F)

represents cards that are both hearts AND face cards—that's the Jack, Queen, and King of hearts, so

P(H \cap F) = \frac{3}{52}

. Therefore:

P(H \cup F) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52}

\frac{13}{52} + \frac{12}{52} = \frac{25}{52}

. Remember: whenever you see "or" in probability (union), always check for overlap between events. The inclusion-exclusion principle prevents double-counting and is essential for union problems.

Question 4

In a city, 60% of households subscribe to a streaming service, 50% subscribe to cable television, and 20% subscribe to neither. What is the probability that a randomly selected household subscribes to both?

0.10
0.20
0.30 (correct answer)
0.40

Explanation: Let S be the event of subscribing to a streaming service and C be the event of subscribing to cable television. We are given

P(S)=0.6

P(C)=0.5

, and

P(S^c \cap C^c)=0.2

. The event 'subscribing to neither' is the complement of the union of S and C. By De Morgan's laws,

P(S^c \cap C^c) = P((S \cup C)^c) = 1 - P(S \cup C)

. So,

0.2 = 1 - P(S \cup C)

, which means

P(S \cup C) = 0.8

. Now we use the general addition rule to find the intersection:

P(S \cup C) = P(S) + P(C) - P(S \cap C)

. Substituting the values gives

0.8 = 0.6 + 0.5 - P(S \cap C)

, which simplifies to

0.8 = 1.1 - P(S \cap C)

. Therefore,

P(S \cap C) = 0.3

Question 5

In a survey of pet owners, 45% own a dog, 30% own a cat, and 15% own both a dog and a cat. What is the probability that a randomly selected pet owner owns neither a dog nor a cat?

0.10
0.25
0.40 (correct answer)
0.55

Explanation: Let D be the event of owning a dog and C be the event of owning a cat. We are given

P(D) = 0.45

P(C) = 0.30

, and

P(D \cap C) = 0.15

. The probability of owning at least one of these pets is the union,

P(D \cup C) = P(D) + P(C) - P(D \cap C) = 0.45 + 0.30 - 0.15 = 0.60

. The probability of owning neither is the complement of this union,

P((D \cup C)^c) = 1 - P(D \cup C) = 1 - 0.60 = 0.40

Question 6

A company has two backup power generators, G1 and G2. The probability that G1 fails is 0.10, the probability that G2 fails is 0.08, and the probability that both fail is 0.03. What is the probability that at least one generator is operational?

0.15
0.85
0.97 (correct answer)
0.992

Explanation: Let F1 be the event that G1 fails, and F2 be the event that G2 fails. We are given

P(F1) = 0.10

P(F2) = 0.08

, and

P(F1 \cap F2) = 0.03

. The event 'at least one generator is operational' is the complement of the event 'both generators fail'. The probability that both generators fail is given as

P(F1 \cap F2) = 0.03

. Therefore, the probability that at least one is operational is

1 - P(F1 \cap F2) = 1 - 0.03 = 0.97

Question 7

Let A, B, and C be three independent events with $P(A) = 0.5$ , $P(B) = 0.4$ , and $P(C) = 0.2$ . What is the probability that at least one of these events occurs?

0.04
0.76 (correct answer)
0.80
1.10

Explanation: The probability that at least one event occurs is the complement of the probability that none of the events occur. The probability of none occurring is

P(A^c \cap B^c \cap C^c)

. Since A, B, and C are independent, so are their complements. First, find the probabilities of the complements:

P(A^c) = 1 - 0.5 = 0.5

P(B^c) = 1 - 0.4 = 0.6

, and

P(C^c) = 1 - 0.2 = 0.8

. Then,

P(A^c \cap B^c \cap C^c) = P(A^c)P(B^c)P(C^c) = 0.5 \times 0.6 \times 0.8 = 0.24

. The probability of at least one event occurring is

1 - 0.24 = 0.76

Question 8

Events E and F are independent, with $P(E) = 0.4$ and $P(F) = 0.5$ . What is the probability of the event $E^c \cup F$ ?

0.90
0.80 (correct answer)
0.70
0.60

Explanation: We use the general addition rule:

P(E^c \cup F) = P(E^c) + P(F) - P(E^c \cap F)

. First, find

P(E^c) = 1 - P(E) = 1 - 0.4 = 0.6

. Because events E and F are independent, their complements are also independent, so

E^c

and F are independent. Thus,

P(E^c \cap F) = P(E^c) \cdot P(F) = 0.6 \cdot 0.5 = 0.3

. Substituting these values into the addition rule gives

P(E^c \cup F) = 0.6 + 0.5 - 0.3 = 0.8

Question 9

For two events, A and B, it is known that $P(A \cup B) = 0.75$ and $P(A^c) = 0.40$ . The events A and B are independent. What is the value of $P(B)$ ?

0.25
0.375 (correct answer)
0.40
0.50

Explanation: First, find

P(A)

from its complement:

P(A) = 1 - P(A^c) = 1 - 0.40 = 0.60

. Let

P(B) = p

. Since A and B are independent,

P(A \cap B) = P(A)P(B) = 0.60p

. Now use the general addition rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Substitute the known values:

0.75 = 0.60 + p - 0.60p

. Simplify the equation:

0.15 = 0.40p

. Solve for p:

p = 0.15 / 0.40 = 0.375

Question 10

Given that $P(X) = 0.7$ and $P(Y) = 0.6$ . What is the minimum possible value for the probability of the intersection $P(X \cap Y)$ ?

0.0
0.3 (correct answer)
0.42
0.6

Explanation: The general addition rule states

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)

. We know that the probability of any event, including the union, cannot exceed 1. So,

P(X \cup Y) \le 1

. Substituting the given probabilities, we have

0.7 + 0.6 - P(X \cap Y) \le 1

, which simplifies to

1.3 - P(X \cap Y) \le 1

. Rearranging this inequality to solve for

P(X \cap Y)

gives

0.3 \le P(X \cap Y)

. Therefore, the minimum possible value for the intersection is 0.3.

Question 11

For two events, A and B, $P(A) = 0.5$ and $P(A \cup B) = 0.8$ . Let $P(B) = p$ . If A and B are mutually exclusive, then $p = p_1$ . If A and B are independent, then $p = p_2$ . What is the value of $p_1 + p_2$ ?

0.6
0.8
0.9 (correct answer)
1.1

Explanation: Case 1 (Mutually Exclusive): The addition rule is

P(A \cup B) = P(A) + P(B)

. Substituting the values,

0.8 = 0.5 + p_1

, which gives

p_1 = 0.3

. Case 2 (Independent): The addition rule is

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Since they are independent,

P(A \cap B) = P(A)P(B) = 0.5 p_2

. Substituting gives

0.8 = 0.5 + p_2 - 0.5 p_2

. This simplifies to

0.3 = 0.5 p_2

, which means

p_2 = 0.6

. The required sum is

p_1 + p_2 = 0.3 + 0.6 = 0.9

Question 12

A manufacturer finds that 10% of its products have a cosmetic flaw and 5% have a functional flaw. If 2% of products have both flaws, what is the probability that a randomly selected product has a functional flaw but no cosmetic flaw?

0.03 (correct answer)
0.05
0.08
0.13

Explanation: Let C be the event of having a cosmetic flaw and F be the event of having a functional flaw. We are given

P(C) = 0.10

P(F) = 0.05

, and

P(C \cap F) = 0.02

. The question asks for the probability of a functional flaw but no cosmetic flaw, which is represented by

P(F \cap C^c)

. The formula for this is

P(F \cap C^c) = P(F) - P(F \cap C)

. Substituting the given values:

0.05 - 0.02 = 0.03

Question 13

Let E and F be events such that $P(E) = 0.4$ and $P(E \cup F) = 0.7$ . If E and F are mutually exclusive, what is $P(F)$ ?

0.30 (correct answer)
0.50
0.60
0.70

Explanation: If two events E and F are mutually exclusive, it means they cannot occur at the same time, so

P(E \cap F) = 0

. The general addition rule

P(E \cup F) = P(E) + P(F) - P(E \cap F)

simplifies to

P(E \cup F) = P(E) + P(F)

. Substituting the given probabilities, we have

0.7 = 0.4 + P(F)

. Solving for

P(F)

gives

P(F) = 0.7 - 0.4 = 0.3

Question 14

If $P(A \cap B) = 0.2$ and $P(A^c \cap B^c) = 0.3$ , what is $P(A) + P(B)$ ?

0.5
0.7
0.9 (correct answer)
1.2

Explanation: We are given

P(A^c \cap B^c) = 0.3

. By De Morgan's laws, this is equal to

P((A \cup B)^c)

. Using the complement rule,

P(A \cup B) = 1 - P((A \cup B)^c) = 1 - 0.3 = 0.7

. The general addition rule is

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. We can rearrange this to find the sum we need:

P(A) + P(B) = P(A \cup B) + P(A \cap B)

. Substituting the values we know:

P(A) + P(B) = 0.7 + 0.2 = 0.9

Question 15

0.6
0.8
0.9 (correct answer)
1.1

Explanation: Case 1 (Mutually Exclusive): The addition rule is

P(A \cup B) = P(A) + P(B)

. Substituting the values,

0.8 = 0.5 + p_1

, which gives

p_1 = 0.3

. Case 2 (Independent): The addition rule is

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Since they are independent,

P(A \cap B) = P(A)P(B) = 0.5 p_2

. Substituting gives

0.8 = 0.5 + p_2 - 0.5 p_2

. This simplifies to

0.3 = 0.5 p_2

, which means

p_2 = 0.6

. The required sum is

p_1 + p_2 = 0.3 + 0.6 = 0.9

Question 16

Given that $P(X) = 0.7$ and $P(Y) = 0.6$ . What is the minimum possible value for the probability of the intersection $P(X \cap Y)$ ?

0.0
0.3 (correct answer)
0.42
0.6

Explanation: The general addition rule states

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)

. We know that the probability of any event, including the union, cannot exceed 1. So,

P(X \cup Y) \le 1

. Substituting the given probabilities, we have

0.7 + 0.6 - P(X \cap Y) \le 1

, which simplifies to

1.3 - P(X \cap Y) \le 1

. Rearranging this inequality to solve for

P(X \cap Y)

gives

0.3 \le P(X \cap Y)

. Therefore, the minimum possible value for the intersection is 0.3.

Question 17

0.03 (correct answer)
0.05
0.08
0.13

Explanation: Let C be the event of having a cosmetic flaw and F be the event of having a functional flaw. We are given

P(C) = 0.10

P(F) = 0.05

, and

P(C \cap F) = 0.02

. The question asks for the probability of a functional flaw but no cosmetic flaw, which is represented by

P(F \cap C^c)

. The formula for this is

P(F \cap C^c) = P(F) - P(F \cap C)

. Substituting the given values:

0.05 - 0.02 = 0.03

Question 18

0.10
0.20
0.30 (correct answer)
0.40

Explanation: Let S be the event of subscribing to a streaming service and C be the event of subscribing to cable television. We are given

P(S)=0.6

P(C)=0.5

, and

P(S^c \cap C^c)=0.2

. The event 'subscribing to neither' is the complement of the union of S and C. By De Morgan's laws,

P(S^c \cap C^c) = P((S \cup C)^c) = 1 - P(S \cup C)

. So,

0.2 = 1 - P(S \cup C)

, which means

P(S \cup C) = 0.8

. Now we use the general addition rule to find the intersection:

P(S \cup C) = P(S) + P(C) - P(S \cap C)

. Substituting the values gives

0.8 = 0.6 + 0.5 - P(S \cap C)

, which simplifies to

0.8 = 1.1 - P(S \cap C)

. Therefore,

P(S \cap C) = 0.3

Question 19

For events E and F, $P(E \cap F) = 0.25$ and $P(E \cup F) = 0.85$ . If the probability of event E is equal to the probability of event F, what is $P(E)$ ?

0.30
0.425
0.50
0.55 (correct answer)

Explanation: Let

x = P(E) = P(F)

. The general addition rule is

P(E \cup F) = P(E) + P(F) - P(E \cap F)

. Substituting the given values and the variable x:

0.85 = x + x - 0.25

. This simplifies to

0.85 = 2x - 0.25

. Adding 0.25 to both sides gives

1.10 = 2x

. Dividing by 2 gives

x = 0.55

. Thus,

P(E) = 0.55

Question 20

Given that $P(A^c) = 0.4$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.8$ . What is $P(A^c \cup B^c)$ ?

0.2
0.3
0.7 (correct answer)
0.9

Explanation: This is a multi-step problem. First, find

P(A) = 1 - P(A^c) = 1 - 0.4 = 0.6

. Next, use the general addition rule to find the intersection:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Substituting the values:

0.8 = 0.6 + 0.5 - P(A \cap B)

, which means

0.8 = 1.1 - P(A \cap B)

, so

P(A \cap B) = 0.3

. The question asks for

P(A^c \cup B^c)

. By De Morgan's laws, this is equal to

P((A \cap B)^c)

. Using the complement rule,

P((A \cap B)^c) = 1 - P(A \cap B) = 1 - 0.3 = 0.7