Basic Probability Rules - College Statistics

Given that $P(X) = 0.7$ and $P(Y) = 0.6$ . What is the minimum possible value for the probability of the intersection $P(X \cap Y)$ ?
Answer: 0.3 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.
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?
Answer: 0.40 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, $P((D \cup C)^c) = 1 - P(D \cup C) = 1 - 0.60 = 0.40$ .
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?
Answer: 0.30 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, $P(S \cap C) = 0.3$ .
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?
Answer: 0.76 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 $1 - 0.24 = 0.76$ .
At a university, 30% of students take a statistics course, 25% take a computer science course, and 10% take both. What percentage of students take exactly one of these two courses?
Answer: 0.35 Let S be the event of taking statistics and C be the event of taking computer science. We are given $P(S) = 0.30$ , $P(C) = 0.25$ , and $P(S \cap C) = 0.10$ . The probability of taking 'exactly one' is the probability of taking statistics but not computer science, plus the probability of taking computer science but not statistics. This is $P(S \cap C^c) + P(C \cap S^c)$ . We calculate these as $P(S \cap C^c) = P(S) - P(S \cap C) = 0.30 - 0.10 = 0.20$ , and $P(C \cap S^c) = P(C) - P(S \cap C) = 0.25 - 0.10 = 0.15$ . The sum is $0.20 + 0.15 = 0.35$ .
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?
Answer: 0.03 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: $0.05 - 0.02 = 0.03$ .
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 $p_1 + p_2$ ?
Answer: 0.9 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 $p_1 + p_2 = 0.3 + 0.6 = 0.9$ .
For two events, A and B, it is known that $P(A) = 0.6$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.8$ . What is the probability that event A occurs and event B does not occur?
Answer: 0.3 The question asks for $P(A \cap B^c)$ . The formula for this is $P(A) - P(A \cap B)$ . To use this, we first need to find $P(A \cap B)$ using the general addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . Substituting the given values: $0.8 = 0.6 + 0.5 - P(A \cap B)$ , which gives $0.8 = 1.1 - P(A \cap B)$ , so $P(A \cap B) = 0.3$ . Now we can find the desired probability: $P(A \cap B^c) = P(A) - P(A \cap B) = 0.6 - 0.3 = 0.3$ .
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?
Answer: 0.97 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 $1 - P(F1 \cap F2) = 1 - 0.03 = 0.97$ .
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 $P(F)$ ?
Answer: 0.30 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 $P(F) = 0.7 - 0.4 = 0.3$ .
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 $P(E)$ ?
Answer: 0.55 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, $P(E) = 0.55$ .
If $P(A \cap B) = 0.2$ and $P(A^c \cap B^c) = 0.3$ , what is $P(A) + P(B)$ ?
Answer: 0.9 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: $P(A) + P(B) = 0.7 + 0.2 = 0.9$ .
Two events, X and Y, are such that $P(X) = 0.5$ , $P(Y) = 0.3$ , and $P(X \cup Y) = 0.65$ . Which of the following best describes the relationship between X and Y?
Answer: They are independent. When you encounter probability questions involving two events and their union, you need to determine the relationship between the events using the fundamental probability formulas. Start with the addition rule: $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$ . Substituting the given values: $0.65 = 0.5 + 0.3 - P(X \cap Y)$ , which gives us $P(X \cap Y) = 0.15$ . Now test for independence by checking if $P(X \cap Y) = P(X) \times P(Y)$ . Calculate: $P(X) \times P(Y) = 0.5 \times 0.3 = 0.15$ . Since $P(X \cap Y) = 0.15$ equals $P(X) \times P(Y) = 0.15$ , the events are independent. This confirms answer D is correct. Let's examine why the other options fail. Choice A (mutually exclusive) would require $P(X \cap Y) = 0$ , but we calculated $P(X \cap Y) = 0.15$ , so they're not mutually exclusive. Choice B (complements) would require $P(X) + P(Y) = 1$ , but $0.5 + 0.3 = 0.8 \neq 1$ , and complements cannot overlap, yet these events do. Choice C incorrectly states they're neither mutually exclusive nor independent—while they're indeed not mutually exclusive, we've proven they are independent. Study tip: Always calculate $P(X \cap Y)$ first using the addition rule, then test specific relationships. For independence, check if $P(X \cap Y) = P(X) \times P(Y)$ . This systematic approach prevents confusion between the different types of event relationships.
Let A and B be two events such that $P(A) > 0$ and $P(B) > 0$ . If A and B are mutually exclusive, which of the following statements must be true?
Answer: The events A and B are not independent. When you encounter questions about mutually exclusive events and independence, you're dealing with two fundamental but distinct concepts in probability that students often confuse. Mutually exclusive events cannot occur simultaneously, meaning $P(A \cap B) = 0$ . Since we're told $P(A) > 0$ and $P(B) > 0$ , let's examine what this means for independence. Two events are independent if $P(A \cap B) = P(A) \cdot P(B)$ . But if A and B are mutually exclusive, then $P(A \cap B) = 0$ . For independence to hold, we'd need $P(A) \cdot P(B) = 0$ , which would require either $P(A) = 0$ or $P(B) = 0$ . Since both probabilities are positive, this is impossible. Therefore, mutually exclusive events with positive probabilities are never independent, making D correct. Option A is wrong because it claims the opposite of what we just proved. Option B is incorrect because mutually exclusive events don't need to be exhaustive—there could be other outcomes besides A and B, so $P(A) + P(B)$ could be less than 1. Option C confuses the union formula with the independence condition. For mutually exclusive events, $P(A \cup B) = P(A) + P(B)$ , not $P(A)P(B)$ . Remember this key insight: mutually exclusive events with positive probabilities are always dependent. This makes intuitive sense—if knowing that A occurred tells you that B definitely didn't occur, then A and B are clearly not independent.
Events E and F are independent, with $P(E) = 0.4$ and $P(F) = 0.5$ . What is the probability of the event $E^c \cup F$ ?
Answer: 0.80 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 $P(E^c \cup F) = 0.6 + 0.5 - 0.3 = 0.8$ .
Given that $P(A^c) = 0.4$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.8$ . What is $P(A^c \cup B^c)$ ?
Answer: 0.7 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, $P((A \cap B)^c) = 1 - P(A \cap B) = 1 - 0.3 = 0.7$ .
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 $P(H \cup F)$ ?
Answer: 22/52 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.
The Venn diagram below shows probabilities for events A, B, and C. What is the value of $P(A \cup (B \cap C))$ ?
Answer: 0.4 The event $A \cup (B \cap C)$ represents the union of event A and the intersection of events B and C. We can find this probability by summing the probabilities of the disjoint regions that make up this event. These regions are: A only, (A and B only), (A and C only), (A and B and C), and (B and C only). Based on the diagram, the sum is $0.10 + 0.08 + 0.12 + 0.03 + 0.07 = 0.40$ . Alternatively, using the formula $P(A \cup (B \cap C)) = P(A) + P(B \cap C) - P(A \cap B \cap C)$ . First, $P(A) = 0.10 + 0.08 + 0.12 + 0.03 = 0.33$ . Second, $P(B \cap C) = 0.07 + 0.03 = 0.10$ . Third, $P(A \cap B \cap C) = 0.03$ . The result is $0.33 + 0.10 - 0.03 = 0.40$ .
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 $P(B)$ ?
Answer: 0.375 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: $p = 0.15 / 0.40 = 0.375$ .
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?
Answer: 0.02 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 $P(F1 \cap F2) = P(F1) \cdot P(F2) = 0.1 \cdot 0.2 = 0.02$ .

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