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formed."},{"text":"sample point","correct":false,"explanation":"A sample point is one individual outcome, not the set of all possible outcomes. The complete set is called the sample space."},{"text":"event","correct":false,"explanation":"An event is a subset of possible outcomes satisfying a condition, whereas the sample space is the set containing every possible outcome."},{"text":"probability distribution","correct":false,"explanation":"A probability distribution assigns probabilities to outcomes or values; it is not the set itself. The set of all outcomes is the sample space."}]},{"id":"7598cbe5-691e-4ca4-ae6a-e28e67767900","cardType":"basic","points":1,"front":"What range of values can a probability $P(A)$ take?","back":"$$0 \\le P(A) \\le 1$","mcqOptions":[{"text":"$$0 \\le P(A) \\le 1$","correct":true,"explanation":"A probability cannot be negative or exceed one. Zero represents an impossible event, while one represents an event that is certain."},{"text":"$$0 < P(A) < 1$","correct":false,"explanation":"Using strict inequalities wrongly excludes impossible and certain events. Probabilities may equal zero or one, so the correct range is $0 \\le P(A) \\le 1$."},{"text":"$$-1 \\le P(A) \\le 1$","correct":false,"explanation":"The range from $-1$ to $1$ applies to quantities such as correlation coefficients, not probabilities. Probabilities lie between $0$ and $1$ inclusive."},{"text":"$$0 \\le P(A) \\le 100$","correct":false,"explanation":"This confuses probabilities with percentages. A probability is expressed from $0$ to $1$; percentages range from $0$ to $100$ after multiplication by $100$."}]},{"id":"4cde314a-63bb-42a7-b01c-9b20ffb3f524","cardType":"cloze","points":null,"front":"For any event $A$, the probability of its complement $A'$ is calculated as $P(A') = \\text{[...]}$.","back":"For any event $A$, the probability of its complement $A'$ is calculated as $P(A') = \\boxed{1 - P(A)}$.","mcqOptions":[{"text":"$$1 - P(A)$","correct":true,"explanation":"An event and its complement together cover the entire sample space and have total probability $1$, so subtracting $P(A)$ gives $P(A')$."},{"text":"$$P(A)$","correct":false,"explanation":"This treats an event and its complement as having equal probabilities, which is generally false. Since their probabilities sum to $1$, $P(A')=1-P(A)$."},{"text":"$$1 + P(A)$","correct":false,"explanation":"Adding $P(A)$ to $1$ produces a value above $1$ whenever $A$ is possible. Complementary probabilities sum to $1$, so $P(A')=1-P(A)$."},{"text":"$$\\frac{1}{P(A)}$","correct":false,"explanation":"Taking a reciprocal is not the complement rule and can produce probabilities greater than $1$. The missing probability is found using $P(A')=1-P(A)$."}]},{"id":"3b459ebe-3bbf-43ca-91bc-b82c7cfaee95","cardType":"basic","points":1,"front":"In set notation, what symbol represents the intersection (A and B)?","back":"$$A \\cap B$","mcqOptions":[{"text":"$$A \\cap B$","correct":true,"explanation":"The intersection $A \\cap B$ contains outcomes belonging to both $A$ and $B$, so it represents the event that A and B occur."},{"text":"$$A \\cup B$","correct":false,"explanation":"The symbol $\\cup$ denotes the union, containing outcomes in A or B or both. The intersection of A and B is written $A \\cap B$."},{"text":"$$A'$","correct":false,"explanation":"The prime denotes the complement of A, containing outcomes not in A. Outcomes common to A and B are represented by $A \\cap B$."},{"text":"$$A \\setminus B$","correct":false,"explanation":"This denotes the set difference, containing outcomes in A but not B. The outcomes shared by both events are written $A \\cap B$."}]},{"id":"6f66d336-523c-42b5-8a8f-2262c2f4c0ec","cardType":"basic","points":1,"front":"In set notation, what symbol represents the union (A or B)?","back":"$$A \\cup B$","mcqOptions":[{"text":"$$A \\cup B$","correct":true,"explanation":"The union $A \\cup B$ contains every outcome belonging to A, B, or both, so it represents the inclusive event “A or B”."},{"text":"$$A \\cap B$","correct":false,"explanation":"The symbol $\\cap$ denotes the intersection, containing only outcomes common to both events. The inclusive event A or B is written $A \\cup B$."},{"text":"$$A'$","correct":false,"explanation":"The prime denotes the complement of A, meaning outcomes outside A. The event containing outcomes in A or B is written $A \\cup B$."},{"text":"$$A \\setminus B$","correct":false,"explanation":"This denotes outcomes in A but not B, so it excludes part of the required union. A or B is represented by $A \\cup B$."}]},{"id":"28b4c312-2c89-4e16-b935-c73adb553f6c","cardType":"basic","points":2,"front":"In a Venn diagram, why is $P(A \\cap B)$ subtracted in the formula for $P(A \\cup B)$?","back":"To avoid double-counting the overlap between the two events.","mcqOptions":[{"text":"To avoid double-counting the overlap between the two events.","correct":true,"explanation":"Adding $P(A)$ and $P(B)$ counts outcomes in both events twice. Subtracting $P(A \\cap B)$ once leaves each overlapping outcome counted exactly once."},{"text":"To make the events mutually exclusive before adding probabilities.","correct":false,"explanation":"Subtracting the overlap does not make the original events mutually exclusive; they may still share outcomes. It corrects the double-counting caused by adding their probabilities."},{"text":"To avoid counting outcomes outside either of the two events.","correct":false,"explanation":"Outcomes outside both events were never included in $P(A)+P(B)$, so they do not need removing. The subtraction corrects double-counting within the overlap."},{"text":"To find the complement of the overlap between the two events.","correct":false,"explanation":"Subtracting the intersection does not produce a complement. It removes one of the two copies of the overlap created when $P(A)$ and $P(B)$ are added."}]},{"id":"426eff3c-b3bd-4724-b8ae-538fe83d0973","cardType":"cloze","points":null,"front":"The addition rule for the union of two events is $P(A \\cup B) = \\text{[...]}$.","back":"The addition rule for the union of two events is $P(A \\cup B) = \\boxed{P(A) + P(B) - P(A \\cap B)}$.","mcqOptions":[{"text":"$$P(A) + P(B) - P(A \\cap B)$","correct":true,"explanation":"Adding $P(A)$ and $P(B)$ counts their intersection twice, so $P(A \\cap B)$ is subtracted once to count every outcome in the union exactly once."},{"text":"$$P(A) + P(B)$","correct":false,"explanation":"Simply adding the probabilities double-counts outcomes in both events. The general rule subtracts the overlap: $P(A \\cup B)=P(A)+P(B)-P(A \\cap B)$."},{"text":"$$P(A) + P(B) + P(A \\cap B)$","correct":false,"explanation":"Adding the intersection counts overlapping outcomes three times rather than once. It must be subtracted, giving $P(A)+P(B)-P(A \\cap B)$."},{"text":"$$P(A) + P(B) - 2P(A \\cap B)$","correct":false,"explanation":"The overlap is initially counted twice, so subtracting it twice removes it completely. Subtracting it once gives $P(A)+P(B)-P(A \\cap B)$."}]},{"id":"fb58a8d7-bb47-4b6a-aeca-da8d594abcc6","cardType":"basic","points":1,"front":"Define mutually exclusive events.","back":"Events that cannot happen at the same time.","mcqOptions":[{"text":"Events that cannot happen at the same time.","correct":true,"explanation":"Mutually exclusive events have no outcomes in common, so they cannot occur together and their intersection has probability zero."},{"text":"Events that always happen at the same time.","correct":false,"explanation":"Events that always occur together have the same occurrence pattern, not mutual exclusivity. Mutually exclusive events cannot happen at the same time."},{"text":"Events that have equal probabilities in the same experiment.","correct":false,"explanation":"Equal probabilities do not prevent events from occurring together. Mutually exclusive events are specifically events that cannot happen at the same time."},{"text":"Events that cover all outcomes in the sample space.","correct":false,"explanation":"Covering all outcomes describes exhaustive events, which is a different property. Mutually exclusive events are those that cannot happen at the same time."}]},{"id":"03ba26e2-17f1-4a01-87e5-2af9adbf0c94","cardType":"cloze","points":null,"front":"If events $A$ and $B$ are mutually exclusive, then $P(A \\cap B) = \\text{[...]}$.","back":"If events $A$ and $B$ are mutually exclusive, then $P(A \\cap B) = \\boxed{0}$.","mcqOptions":[{"text":"0","correct":true,"explanation":"Mutually exclusive events cannot occur together, so their intersection is empty and its probability is $0$."},{"text":"1","correct":false,"explanation":"A probability of $1$ would mean the events are certain to occur together. Mutually exclusive events cannot occur together, so $P(A \\cap B)=0$."},{"text":"$$P(A)$","correct":false,"explanation":"This would suggest that whenever $A$ occurs, $B$ also occurs. Mutually exclusive events have no shared outcomes, so $P(A \\cap B)=0$."},{"text":"$$P(B)$","correct":false,"explanation":"This would suggest that whenever $B$ occurs, $A$ also occurs. Mutually exclusive events cannot occur simultaneously, so $P(A \\cap B)=0$."}]},{"id":"4544c671-ea9b-4e36-b5b9-5d68f62c2ae1","cardType":"basic","points":2,"front":"If two events are mutually exclusive and have non-zero probabilities, can they be independent?","back":"No; knowing one occurs makes the other impossible, so they affect each other.","mcqOptions":[{"text":"No; knowing one occurs makes the other impossible, so they affect each other.","correct":true,"explanation":"For non-zero mutually exclusive events, $P(A \\cap B)=0$ but $P(A)P(B)>0$. Therefore the independence condition is not satisfied."},{"text":"Yes; having no outcomes in common means that they do not affect each other.","correct":false,"explanation":"This confuses mutually exclusive with independent. If one mutually exclusive event occurs, the other becomes impossible, so non-zero mutually exclusive events are not independent."},{"text":"Yes; knowing one occurs leaves the probability of the other event unchanged.","correct":false,"explanation":"The probability does change: after one event occurs, the conditional probability of the other is $0$. Therefore non-zero mutually exclusive events cannot be independent."},{"text":"No; independent events must always occur together, so they cannot be mutually exclusive.","correct":false,"explanation":"Independent events do not have to occur together; they satisfy $P(A \\cap B)=P(A)P(B)$. Non-zero mutually exclusive events fail this because their intersection has probability $0$."}]},{"id":"0e4bd064-46a5-4703-a2fc-e9147a0858c6","cardType":"basic","points":1,"front":"What is the first step when filling in a three-set Venn diagram?","back":"Place the value for the center region (all three events) first.","mcqOptions":[{"text":"Place the value for the center region (all three events) first.","correct":true,"explanation":"The all-three-events intersection contributes to every pairwise intersection. Filling it first allows that shared amount to be subtracted correctly when calculating the surrounding regions."},{"text":"Place the values for the outer regions (one event only) first.","correct":false,"explanation":"The one-event-only regions depend on values already assigned to overlapping regions. Start with the center region, where all three events intersect, and work outwards."},{"text":"Place the value for the outside region (none of the events) first.","correct":false,"explanation":"The outside region is usually found after the event regions have been completed. Begin with the center region representing the intersection of all three events."},{"text":"Place the values for the pairwise regions (two events only) first.","correct":false,"explanation":"Given pairwise intersections often include the center value, so entering them first can cause double-counting. Place the all-three-events center region first, then subtract it where necessary."}]},{"id":"627b56bb-61b5-4e7d-a28f-b34d0fdb53fa","cardType":"basic","points":1,"front":"Define independent events.","back":"The outcome of one event does not affect the probability of the other.","mcqOptions":[{"text":"The outcome of one event does not affect the probability of the other.","correct":true,"explanation":"Independence means that knowing whether one event occurred provides no information that changes the probability of the other event."},{"text":"The outcome of one event makes impossible the occurrence of the other.","correct":false,"explanation":"This describes mutually exclusive events, not independent events. For independent events, the outcome of one does not affect the probability of the other."},{"text":"The outcome of one event increases the probability of the other.","correct":false,"explanation":"An increased probability indicates dependence because information about one event changes the likelihood of the other. Independent events leave the other probability unchanged."},{"text":"The outcome of one event determines completely the outcome of the other.","correct":false,"explanation":"If one outcome determines the other, the events are strongly dependent. Independent events are defined by one outcome not affecting the probability of the other."}]},{"id":"6525d840-acc4-4147-9cb7-f86ddfa4d546","cardType":"cloze","points":null,"front":"Events $A$ and $B$ are independent if and only if $P(A \\cap B) = \\text{[...]}$.","back":"Events $A$ and $B$ are independent if and only if $P(A \\cap B) = \\boxed{P(A)P(B)}$.","mcqOptions":[{"text":"$$P(A)P(B)$","correct":true,"explanation":"For independent events, the probability that both occur equals the product of their individual probabilities: $P(A \\cap B)=P(A)P(B)$."},{"text":"$$P(A)+P(B)$","correct":false,"explanation":"Adding individual probabilities is associated with unions, with an overlap adjustment where necessary. Independence requires the intersection to equal $P(A)P(B)$."},{"text":"$$P(A)+P(B)-1$","correct":false,"explanation":"This expression can arise as a lower bound for an intersection, but it is not the independence condition. Independent events satisfy $P(A \\cap B)=P(A)P(B)$."},{"text":"$$P(A)-P(B)$","correct":false,"explanation":"Subtracting the event probabilities does not calculate their intersection or test independence. The required condition is $P(A \\cap B)=P(A)P(B)$."}]},{"id":"b6cfee48-238d-428e-9ea3-c9b0746fef61","cardType":"basic","points":2,"front":"When asked to test for independence in an exam, what must your answer include?","back":"A comparison showing whether $P(A \\cap B)$ equals $P(A)P(B)$.","mcqOptions":[{"text":"A comparison showing whether $P(A \\cap B)$ equals $P(A)P(B)$.","correct":true,"explanation":"Independence is tested using $P(A \\cap B)=P(A)P(B)$. The answer must calculate or state both sides and explicitly compare them."},{"text":"A comparison showing whether $P(A \\cup B)$ equals $P(A)+P(B)$.","correct":false,"explanation":"This equality tests whether the events are mutually exclusive, since it assumes no overlap. Independence must be tested by comparing $P(A \\cap B)$ with $P(A)P(B)$."},{"text":"A comparison showing whether $P(A \\cap B)$ equals $P(A)+P(B)$.","correct":false,"explanation":"This incorrectly adds probabilities when checking an intersection. To test independence, compare the intersection probability $P(A \\cap B)$ with the product $P(A)P(B)$."},{"text":"A comparison showing whether $P(A \\cup B)$ equals $P(A)P(B)$.","correct":false,"explanation":"The product rule for independence concerns both events occurring, represented by an intersection, not a union. Compare $P(A \\cap B)$ with $P(A)P(B)$."}]},{"id":"2c507209-94e7-4cda-8dee-bd0c30b27677","cardType":"basic","points":1,"front":"In a tree diagram, how do you calculate the probability of a specific combined outcome (e.g., A and then B)?","back":"Multiply the probabilities along the branches.","mcqOptions":[{"text":"Multiply the probabilities along the branches.","correct":true,"explanation":"A specific route represents successive outcomes occurring together, so the multiplication rule is used to find the probability of following that complete route."},{"text":"Add the probabilities along the branches.","correct":false,"explanation":"Adding is used to combine probabilities from different qualifying routes. For successive outcomes along one specific route, multiply the branch probabilities."},{"text":"Subtract the second probability from the first.","correct":false,"explanation":"Subtraction does not represent successive outcomes on a tree. A particular combined outcome is found by multiplying the probabilities along its route."},{"text":"Divide the first probability by the second.","correct":false,"explanation":"Division may appear when calculating conditional probabilities, but it does not give a route probability. Multiply the probabilities along the relevant branches."}]},{"id":"64a193c5-bf4a-4eed-a00f-e86e8ea1d5ce","cardType":"cloze","points":null,"front":"When multiple different routes on a tree diagram satisfy a condition, you [ ] their individual probabilities.碎","back":"When multiple different routes on a tree diagram satisfy a condition, you add their individual probabilities.碎","mcqOptions":[{"text":"add","correct":true,"explanation":"Different routes represent mutually exclusive ways for the required condition to occur. Their separate route probabilities are therefore added to obtain the total probability."},{"text":"multiply","correct":false,"explanation":"Multiplication is used along the branches of one route, not between alternative routes. When several different routes satisfy the condition, add their probabilities."},{"text":"subtract","correct":false,"explanation":"Subtracting route probabilities would remove one valid way of satisfying the condition. Alternative qualifying routes are mutually exclusive, so their probabilities must be added."},{"text":"divide","correct":false,"explanation":"Division is not used to combine alternative outcomes on a tree diagram. Calculate each route probability, then add the probabilities of all routes satisfying the condition."}]},{"id":"36509c19-31ce-47c0-a864-a7f08304024f","cardType":"basic","points":1,"front":"What does the notation $P(B|A)$ represent?","back":"The conditional probability of $B$ given that $A$ has occurred.","mcqOptions":[{"text":"The conditional probability of $B$ given that $A$ has occurred.","correct":true,"explanation":"Conditional probability measures the chance that $B$ occurs when it is already known that $A$ has occurred, so the relevant sample space is restricted to $A$."},{"text":"The conditional probability of $A$ given that $B$ has occurred.","correct":false,"explanation":"This reverses the conditioning: it describes $P(A|B)$, not $P(B|A)$. The correct probability is that of $B$ given that $A$ has occurred."},{"text":"The joint probability of $A$ and $B$ both occurring.","correct":false,"explanation":"A joint probability is written $P(A\\cap B)$ and measures both events occurring. $P(B|A)$ instead measures the probability of $B$ given that $A$ has occurred."},{"text":"The combined probability of $A$ or $B$ occurring.","correct":false,"explanation":"The probability of at least one event occurring is a union, written $P(A\\cup B)$. The notation $P(B|A)$ is the conditional probability of $B$ given $A$."}]},{"id":"3f3c824a-9f2d-43f5-b178-c87858a891cd","cardType":"basic","points":2,"front":"How does \"without replacement\" affect a tree diagram for a second selection?","back":"The total outcomes and the favourable outcomes must both be updated (decremented).","mcqOptions":[{"text":"The total outcomes and the favourable outcomes must both be updated (decremented).","correct":true,"explanation":"After an item is removed, there is one fewer item in total and one fewer item of the selected type, so both counts used on subsequent branches must change."},{"text":"The total outcomes stay unchanged, while the favourable outcomes must be updated (decremented).","correct":false,"explanation":"Keeping the denominator unchanged treats the first item as if it were returned. Without replacement, both the total and favourable outcome counts must be decremented for the second selection."},{"text":"The total outcomes must be updated (decremented), while the favourable outcomes stay unchanged.","correct":false,"explanation":"Changing only the denominator overlooks the removal of an item from its category. Without replacement, both the total outcomes and the favourable outcomes must be updated."},{"text":"The total outcomes and the favourable outcomes must both remain unchanged.","correct":false,"explanation":"Leaving both counts unchanged describes sampling with replacement, where the original composition is restored. Without replacement, both the total and favourable outcome counts must be decremented."}]},{"id":"4acd5635-7c26-4c98-a371-0fd98db39e7e","cardType":"basic","points":2,"front":"When calculating the probability of picking \"one of each\" type from a tree, which routes must be considered?","back":"All possible orders (e.g., Red then Blue AND Blue then Red).","mcqOptions":[{"text":"All possible orders (e.g., Red then Blue AND Blue then Red).","correct":true,"explanation":"Getting one of each type can happen in either order. These routes are mutually exclusive, so their branch products must both be calculated and then added."},{"text":"Only the stated order (e.g., Red then Blue but NOT Blue then Red).","correct":false,"explanation":"Considering only one order omits another valid way to obtain one of each. Both Red then Blue and Blue then Red routes must be included."},{"text":"All matching-type orders (e.g., Red then Red AND Blue then Blue).","correct":false,"explanation":"These routes produce two items of the same type, not one of each. The required routes are Red then Blue and Blue then Red."},{"text":"Only the more likely order (e.g., whichever route has the larger product).","correct":false,"explanation":"Even a less likely order still contributes to the event’s probability. Every route producing one of each must be included, regardless of the sizes of its probabilities."}]},{"id":"ec35e3d4-875e-480a-a129-ee791ba36a4d","cardType":"basic","points":2,"front":"How do you find the total probability of an event that can occur via different machines or categories?","back":"Sum the products of the branches for every route leading to that event.","mcqOptions":[{"text":"Sum the products of the branches for every route leading to that event.","correct":true,"explanation":"Each complete route has probability equal to the product of its branch probabilities. Because the event may arise through mutually exclusive routes, those route probabilities are then summed."},{"text":"Sum the individual branches for every route leading to that event.","correct":false,"explanation":"Adding branches within a route ignores that successive stages must both occur. Multiply along each complete route, then sum the products for every route leading to the event."},{"text":"Multiply the products of the branches for every route leading to that event.","correct":false,"explanation":"Multiplying different complete routes treats them as events that must all occur together. Alternative routes are mutually exclusive, so their branch products must be added instead."},{"text":"Use the product of the branches for the most likely route leading to that event.","correct":false,"explanation":"Using only the most likely route omits other ways the event can occur and underestimates its probability. Sum the branch products for every route leading to the event."}]}]}],"examCountdown":{"examDate":"2027-05-13","paperLabel":"Paper 1"},"heading":"Probability","title":"Probability Flashcards","topbar":{"backHref":"/as-level/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"0ca12138-c2fc-42de-a9b4-c932c8752ef9","walled":"$L39"}],"$L3a","$L3b"]