Sample Spaces & Events - College Statistics

A safety system has three components: A, B, and C. Each component can either work (W) or fail (F). The state of the system can be represented by an ordered triple, e.g., (W, F, W) means A works, B fails, C works. The entire system works if component A works, AND at least one of B or C works. How many outcomes in the sample space correspond to the event that the entire system works?
Answer: 3 The sample space has $2^3 = 8$ possible outcomes. We need to find the subset of outcomes where the system works. The condition is 'A works' AND ('B works' OR 'C works'). First, let's list all outcomes where component A works: \{(W,W,W), (W,W,F), (W,F,W), (W,F,F)\}. Now, from this subset, we apply the second condition: 'at least one of B or C works'. * (W,W,W): B and C work. Satisfies the condition. * (W,W,F): B works. Satisfies the condition. * (W,F,W): C works. Satisfies the condition. * (W,F,F): Neither B nor C works. Does not satisfy the condition. Therefore, there are 3 outcomes where the system works: {(W,W,W), (W,W,F), (W,F,W)}.
An experiment consists of rolling two distinct six-sided dice. Let A be the event that the sum of the dice is a prime number. Let B be the event that both dice show a value greater than 3. Let C be the event that the absolute difference between the values on the dice is 2. Which statement accurately describes the relationship between these events?
Answer: Events A and C are mutually exclusive. Two events are mutually exclusive if they have no outcomes in common (their intersection is empty). Let's check each pair: * A and B: Event A (sum is prime) includes outcomes like (5,6) and (6,5) where the sum is 11. Event B (both dice > 3) also includes (5,6) and (6,5). Since there's an overlap, A and B are not mutually exclusive. * B and C: Event B (both dice > 3) includes outcomes like (4,6) and (6,4). Event C (difference is 2) also includes (4,6) and (6,4). Since there's an overlap, B and C are not mutually exclusive. * A and C: Outcomes in C are {(1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4)}. Their sums are 4, 4, 6, 6, 8, 8, 10, 10, respectively. None of these sums are prime numbers. Therefore, there are no outcomes in common between A and C, so they are mutually exclusive.
The sample space S for an experiment is the set of all ordered pairs $(x, y)$ of integers where $1 \le x \le 3$ and $1 \le y \le 3$ . Let event A be $\{(x, y) | x = y\}$ and event B be $\{(x, y) | y > 1\}$ . How many outcomes are in the event $A^c \cap B$ ?
Answer: 4 First, let's list the elements of the sample space and the events. * The sample space S has $3 \times 3 = 9$ outcomes: {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}. * Event A (x=y) is {(1,1), (2,2), (3,3)}. * Event B (y>1) is {(1,2), (1,3), (2,2), (2,3), (3,2), (3,3)}. We are looking for $A^c \cap B$ . This means we want outcomes that are in B AND not in A. 1. Start with the outcomes in B: {(1,2), (1,3), (2,2), (2,3), (3,2), (3,3)}. 2. Remove any of these outcomes that are also in A. The outcomes in A are {(1,1), (2,2), (3,3)}. The outcomes (2,2) and (3,3) are in both A and B. 3. Removing (2,2) and (3,3) from B leaves us with {(1,2), (1,3), (2,3), (3,2)}. There are 4 outcomes in the event $A^c \cap B$ .
A point $(x, y)$ is chosen at random from the sample space S, which is a square region defined by $0 \le x \le 2$ and $0 \le y \le 2$ . Let A be the event that $x+y < 2$ , and let B be the event that $y > x$ . The event $A \cap B$ corresponds to a region in the x-y plane. Which of the following describes this region?
Answer: A triangle with vertices at (0, 0), (0, 2), and (1, 1). When you encounter problems involving geometric probability, you need to visualize the constraints as regions in the coordinate plane and find their intersection. Let's identify each region systematically. The sample space S is the square with $0 \le x \le 2$ and $0 \le y \le 2$ . Event A requires $x + y < 2$ , which represents all points below the line $x + y = 2$ . This line passes through (0,2) and (2,0), so event A is the triangular region below this line within the square. Event B requires $y > x$ , which represents all points above the line $y = x$ (the diagonal from (0,0) to (2,2)). The intersection $A \cap B$ consists of points that satisfy both conditions: they must be below the line $x + y = 2$ AND above the line $y = x$ . These two lines intersect where $x = x + y = 2$ , giving us the point (1,1). The region $A \cap B$ forms a triangle bounded by these two lines and the y-axis, with vertices at (0,0), (0,2), and (1,1). Choice A incorrectly uses (2,0) as a vertex, which violates $y > x$ . Choice B describes a region that's entirely above the line $x + y = 2$ , violating event A. Choice C suggests a square region that extends beyond both constraint lines. Choice D correctly identifies the triangle with vertices (0,0), (0,2), and (1,1). Study tip: When working with geometric probability, always sketch the constraints on a coordinate plane first. Find intersection points of boundary lines to identify vertices of the resulting region.
There are two boxes. Box 1 contains 1 red and 2 blue marbles. Box 2 contains 2 red and 1 blue marble. An experiment consists of first choosing a box at random, and then drawing one marble from the chosen box. An outcome is an ordered pair (Box number, Marble color). Which set represents the event 'a red marble is drawn'?
Answer: {(1, R), (2, R)} The sample space consists of all possible outcomes, which are pairs indicating the box chosen and the color of the marble drawn. The possible outcomes are: drawing a red from Box 1, (1, R); drawing a blue from Box 1, (1, B); drawing a red from Box 2, (2, R); drawing a blue from Box 2, (2, B). The full sample space is S = {(1, R), (1, B), (2, R), (2, B)}. The event 'a red marble is drawn' is the subset of the sample space containing all outcomes where the marble color is R. This corresponds to the set {(1, R), (2, R)}.
Let S be the sample space of all students at a university. Let A be the event that a student is a math major, and B be the event that a student is in the university choir. Which of the following is the correct verbal description for the event $A \setminus B$ ?
Answer: A student is a math major who is not in the university choir. The notation $A \setminus B$ represents the set difference, which contains all elements that are in set A but not in set B. This can also be written as $A \cap B^c$ . In this context, A is the set of math majors and B is the set of students in the choir. So $A \setminus B$ is the set of students who are math majors AND are not in the choir. This corresponds to the description 'A student is a math major who is not in the university choir.'
In a survey of 200 smartphone users, 120 reported using app A, 90 reported using app B, and 40 reported using both. An outcome in the sample space is a randomly selected user. Consider the event E that a selected user uses neither app A nor app B. How many outcomes are in event E?
Answer: 30 We want to find the number of users in the complement of the union of A and B, which is $|(A \cup B)^c|$ . First, we find the number of users who use at least one of the apps, $|A \cup B|$ , using the Principle of Inclusion-Exclusion: $|A \cup B| = |A| + |B| - |A \cap B|$ . Plugging in the given values: $|A \cup B| = 120 + 90 - 40 = 170$ . This is the number of users who use app A or app B or both. The number of users who use neither app is the total number of users minus the number who use at least one app: Total users - $|A \cup B| = 200 - 170 = 30$ .
An experiment consists of rolling three distinct, fair six-sided dice. Let event A be that the sum of the numbers rolled is less than or equal to 5. How many outcomes are in event A?
Answer: 10 The sample space consists of all ordered triples (d1, d2, d3) where each di is in {1, 2, 3, 4, 5, 6}. The total size is 6³ = 216. We need to find the number of outcomes where the sum is 3, 4, or 5. * Sum = 3: The only possible outcome is (1, 1, 1). (1 outcome) * Sum = 4: The possible outcomes are permutations of (1, 1, 2). These are (1, 1, 2), (1, 2, 1), and (2, 1, 1). (3 outcomes) * Sum = 5: The possible outcomes are permutations of (1, 1, 3) which are (1, 1, 3), (1, 3, 1), (3, 1, 1) and permutations of (1, 2, 2) which are (1, 2, 2), (2, 1, 2), (2, 2, 1). (3 + 3 = 6 outcomes) The total number of outcomes in event A is 1 + 3 + 6 = 10.
A car dealership tracks customer transactions. Let A be the event that a customer purchases a sedan. Let B be the event that a customer purchases a car with a manual transmission. Which of the following describes the event $(A \cup B)^c$ ?
Answer: The customer purchases a vehicle that is not a sedan and does not have a manual transmission. The event $A \cup B$ represents a customer purchasing a sedan or a car with a manual transmission (or both). The complement, $(A \cup B)^c$ , represents all outcomes not in $A \cup B$ . By De Morgan's laws, $(A \cup B)^c = A^c \cap B^c$ . In this context, $A^c$ is the event that the customer does not purchase a sedan, and $B^c$ is the event that the customer does not purchase a car with a manual transmission. The intersection ( $\cap$ ) corresponds to 'and'. Therefore, $A^c \cap B^c$ is the event that the customer purchases a vehicle that is not a sedan AND does not have a manual transmission.
A spinner is divided into three sectors: Red (R), Blue (B), and Green (G). The spinner is spun twice. The sample space S contains outcomes like (R, B), representing Red on the first spin and Blue on the second. Let E be the event that at least one spin is Green. Let F be the event that the two spins are not the same color. Which set of outcomes represents the event $E \cap F$ ?
Answer: {(G, R), (G, B), (R, G), (B, G)} First, let's list the outcomes for events E and F from the sample space S of 9 outcomes. * Event E (at least one Green): E = {(R, G), (B, G), (G, R), (G, B), (G, G)}. * Event F (spins are not the same color): F = {(R, B), (B, R), (R, G), (G, R), (B, G), (G, B)}. The event $E \cap F$ is the intersection of E and F, which includes all outcomes that are in both sets. We need outcomes that have at least one Green AND have different colors on the two spins. Looking at the elements of E, we exclude any where the colors are the same. This removes (G, G). The remaining outcomes in E are {(R, G), (B, G), (G, R), (G, B)}. All of these outcomes consist of two different colors, so they are all in F. Therefore, $E \cap F$ = {(G, R), (G, B), (R, G), (B, G)}.
An urn contains two identical red balls and one blue ball. An experiment consists of drawing two balls, one after another, from the urn without replacement. Which of the following is the correct sample space S for the sequence of colors drawn?
Answer: S = {(R, R), (R, B), (B, R)} When working with sample spaces in probability, you need to identify all possible outcomes of an experiment, considering whether order matters and how to handle identical objects. In this experiment, you're drawing two balls sequentially without replacement from an urn containing two red balls and one blue ball. Since the balls are drawn "one after another," order matters - drawing red then blue is different from drawing blue then red. However, since the two red balls are identical, you cannot distinguish between them in the outcomes. The correct sample space is D: {(R, R), (R, B), (B, R)}. These represent the three possible sequences: red-then-red, red-then-blue, and blue-then-red. This captures all distinguishable outcomes when order matters but the red balls are indistinguishable. A uses set notation {R, R}, {R, B} instead of ordered pairs, which incorrectly suggests order doesn't matter. Additionally, it's missing the (B, R) outcome. B treats the red balls as distinguishable (R1, R2), but the problem states they are identical. This creates an artificially large sample space with six outcomes instead of the three actually distinguishable ones. C only includes (R, B) and (B, R), completely omitting the possibility of drawing both red balls, which is clearly possible since there are two red balls in the urn. Study tip: For sample spaces involving identical objects, focus on what outcomes are actually distinguishable to an observer. Identical objects mean you cannot tell them apart, so don't create separate labels for them in your sample space.
Let S be the sample space of all integers from 1 to 10, inclusive. Which of the following collections of events forms a partition of S?
Answer: E1 = {1, 4, 9}, E2 = {2, 3, 5, 7}, E3 = {6, 8, 10} A collection of events forms a partition of a sample space S if two conditions are met: (1) the events are pairwise mutually exclusive (disjoint), and (2) their union is equal to the entire sample space S. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. * A: E1={2,3,5,7}, E2={4,6,8,9,10}. The union is {2,3,4,5,6,7,8,9,10}. This omits the number 1 (which is neither prime nor composite). The union is not S. * B: E1={1,3,5,7,9}, E2={4,8}. The events are disjoint, but their union is {1,3,4,5,7,8,9}. This omits {2,6,10}. The union is not S. * C: E1={1,4,9}, E2={2,3,5,7}, E3={6,8,10}. These three sets are pairwise disjoint. Their union is {1,2,3,4,5,6,7,8,9,10}, which is S. This collection forms a partition. * D: E1={1,2,3,4,5}, E2={5,6,7,8,9,10}. These events are not mutually exclusive because they both contain the integer 5.
A quality inspector tests light bulbs from a production line one by one. Bulbs are either 'S' (success, it works) or 'F' (failure). The inspector stops as soon as the first failure is found. The sample space for this experiment is $S = \{F, SF, SSF, SSSF, ...\}$ . Let E be the event that the inspector tests at most three bulbs. Which set represents the outcomes in event E?
Answer: {F, SF, SSF} When you encounter a problem about sequential trials that stop at a specific condition, you're dealing with a stopping time experiment. Here, the inspector stops as soon as they find the first failure, so every possible outcome must end with 'F'. The key phrase is "at most three bulbs." This means the inspector tests 1, 2, or 3 bulbs before stopping. Let's trace through each possibility: - Testing exactly 1 bulb: The first bulb fails → outcome is F - Testing exactly 2 bulbs: First succeeds, second fails → outcome is SF - Testing exactly 3 bulbs: First two succeed, third fails → outcome is SSF Any sequence longer than three bulbs (like SSSF) means the inspector tested more than three bulbs, which violates the "at most three" condition. Looking at the wrong answers: Choice A only includes {SSF}, missing the cases where testing stops after 1 or 2 bulbs. Choice B includes {F, SF, SSF, SSSF}, but SSSF represents testing 4 bulbs, which exceeds our limit. Choice C includes {SSS, SSF, SFS, FSS}, but SSS doesn't end in failure (violating the stopping rule), and SFS and FSS are impossible since testing stops immediately after the first failure. Choice D correctly identifies {F, SF, SSF} as the complete set of outcomes where testing stops within three trials. Study tip: In stopping time problems, carefully count the number of trials in each outcome and ensure every sequence follows the stated stopping rule.
A box contains 5 microchips, 2 of which are defective (D) and 3 are not (N). An inspector draws two chips from the box without replacement. The outcome is the ordered pair of results (e.g., (D, N)). Let E be the event that the second chip drawn is defective. How many outcomes are in event E?
Answer: 8 To correctly count the outcomes, it's helpful to imagine the chips are distinct, e.g., D1, D2, N1, N2, N3. The total number of ordered pairs (outcomes) is P(5,2) = 5 * 4 = 20. We are interested in the event E where the second chip is defective. This can happen in two ways: 1. The first chip is defective, and the second is defective (D, D). 2. The first chip is not defective, and the second is defective (N, D). Let's count the number of distinct ways each can happen using the labeled chips: * Case 1 (D, D): There are 2 defective chips (D1, D2). The number of ordered pairs is P(2,2) = 2. The outcomes are (D1, D2) and (D2, D1). * Case 2 (N, D): There are 3 non-defective chips and 2 defective chips. The number of ways to choose one N first is 3. The number of ways to choose one D second is 2. So, there are $3 \times 2 = 6$ outcomes. These are (N1,D1), (N1,D2), (N2,D1), (N2,D2), (N3,D1), (N3,D2). The total number of outcomes in event E is the sum of the outcomes from these two mutually exclusive cases: $2 + 6 = 8$ .
A researcher is studying a population of birds. Let event A be 'the bird is a juvenile' and event B be 'the bird is male'. The researcher observes that events A and $B^c$ are mutually exclusive. What can be concluded from this observation?
Answer: All juvenile birds in the population are male. The event $B^c$ is 'the bird is not male', which means 'the bird is female'. The statement is that events A ('juvenile') and $B^c$ ('female') are mutually exclusive. This means that there are no outcomes where a bird is both a juvenile and a female; i.e., the intersection $A \cap B^c$ is empty. If a bird cannot be both juvenile and female, then any bird that is a juvenile must not be a female. Assuming birds are either male or female, this means all juvenile birds must be male. This implies that event A is a subset of event B ( $A \subseteq B$ ).
A committee of 4 people is to be selected from a group of 10 people consisting of 5 engineers and 5 economists. What is the number of possible committees that contain at least one engineer?
Answer: 205 This problem is best solved using the complement rule. The event 'at least one engineer' is the complement of the event 'zero engineers'. 1. First, find the total number of possible committees of 4 from 10 people, which is given by the combination formula: $\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$ . 2. Next, find the number of committees with 'zero engineers'. This means the committee must be formed entirely of economists. The number of ways to choose 4 economists from 5 is $\binom{5}{4} = \frac{5!}{4!1!} = 5$ . 3. Finally, subtract the number of committees with zero engineers from the total number of committees: $210 - 5 = 205$ .
An experiment consists of flipping a fair coin. If the coin is heads, a four-sided die is rolled. If the coin is tails, a six-sided die is rolled. An outcome is a pair (coin face, die roll). Let E be the event that the die roll is a number greater than 3. How many outcomes are in event E?
Answer: 4 First, we must determine the entire sample space S by considering both cases for the coin flip. * Case 1 (Coin is Heads): The four-sided die can result in {1, 2, 3, 4}. The outcomes are (H, 1), (H, 2), (H, 3), (H, 4). * Case 2 (Coin is Tails): The six-sided die can result in {1, 2, 3, 4, 5, 6}. The outcomes are (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6). The full sample space has 4 + 6 = 10 outcomes. Now, we identify the subset of outcomes for event E, where the die roll is greater than 3. * From Case 1, the roll must be 4. This gives the outcome (H, 4). * From Case 2, the roll can be 4, 5, or 6. This gives the outcomes (T, 4), (T, 5), (T, 6). Combining these, the event E consists of the outcomes {(H, 4), (T, 4), (T, 5), (T, 6)}. There are 4 outcomes in event E.
Consider an experiment where three fair coins are tossed. Let A be the event that there is exactly one head, and let B be the event that there is at least one head. Which statement correctly describes the relationship between these two events?
Answer: A is a proper subset of B. The sample space S for tossing three coins is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. * Event A (exactly one head) is the set {HTT, THT, TTH}. * Event B (at least one head) is the set of all outcomes except for TTT. So, B = {HHH, HHT, HTH, THH, HTT, THT, TTH}. Now we analyze the relationship: * Mutually exclusive? No, because A and B share the outcomes {HTT, THT, TTH}. Their intersection is A itself. * Complement? The complement of B, $B^c$ , is the event of having no heads, which is {TTT}. This is not equal to A. * B is a proper subset of A? No, B contains outcomes like HHH that are not in A. * A is a proper subset of B? Yes. Every outcome in A ({HTT, THT, TTH}) is also an outcome in B. Since B contains other outcomes that are not in A (like HHH), A is a proper subset of B ( $A \subset B$ ).
Two distinct integers are selected simultaneously from the set $\{1, 2, 3, 4, 5, 6\}$ . The outcome of the experiment is the set of the two numbers selected. What is the total number of outcomes in the sample space?
Answer: 15 The problem asks for the number of ways to select two distinct integers from a set of six. The word 'simultaneously' and the fact that the outcome is a 'set' indicate that the order of selection does not matter. This is a combination problem. We need to calculate the number of combinations of choosing 2 items from a set of 6, which is denoted as $\binom{6}{2}$ or C(6, 2). The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ . So, $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$ . There are 15 possible outcomes in the sample space.

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