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possibilities."},{"text":"The complete list of all favourable outcomes.","correct":false,"explanation":"This confuses the sample space with the favourable outcomes for an event. The sample space is the complete list of all possible outcomes, whether favourable or not."},{"text":"The probability of every possible outcome.","correct":false,"explanation":"Probabilities may be assigned after the outcomes are identified, but they are not the sample space itself. The sample space is the complete list of all possible outcomes."},{"text":"The total number of all possible outcomes.","correct":false,"explanation":"This gives the size of the sample space rather than the sample space itself. The sample space is the complete list of the actual possible outcomes."}]},{"id":"6f85b450-2a83-22ba-87c0-80c57438a6cc","cardType":"basic","points":2,"front":"What is the main rule for systematic listing?","back":"Keep one choice fixed, list everything that goes with it, then move to the next choice.","mcqOptions":[{"text":"Keep one choice fixed, list everything that goes with it, then move to the next choice.","correct":true,"explanation":"Fixing one choice while listing every possible partner creates an organised pattern. Repeating this for each choice ensures that no combinations are missed or repeated."},{"text":"Keep one choice fixed, list one thing that goes with it, then move to the next choice.","correct":false,"explanation":"Listing only one partner for each fixed choice omits valid combinations. The correct method lists everything that goes with one fixed choice before moving on."},{"text":"Keep changing every choice, list each result from memory, then check them at the end.","correct":false,"explanation":"Changing all choices at once makes omissions and repetitions more likely. Systematic listing keeps one choice fixed and exhausts its combinations before moving to the next."},{"text":"Keep the choices unordered, list the easiest combinations first, then add any that were missed.","correct":false,"explanation":"Listing easy combinations first does not guarantee completeness and often causes duplicates. The reliable rule is to fix one choice and list everything paired with it."}]},{"id":"d3cc1ef4-6d47-93e7-b441-b77c6dfecf9a","cardType":"cloze","points":null,"front":"An [ ] is one possible result, such as getting heads when you toss a coin.","back":"An outcome is one possible result, such as getting heads when you toss a coin.","mcqOptions":[{"text":"outcome","correct":true,"explanation":"An outcome is one individual result of a probability experiment, such as heads from one coin toss or a six from rolling a die."},{"text":"event","correct":false,"explanation":"An event is a set of one or more outcomes, not necessarily one individual result. The term for one possible result is outcome."},{"text":"sample space","correct":false,"explanation":"A sample space contains every possible outcome of an experiment, rather than describing one result. One possible result is called an outcome."},{"text":"probability","correct":false,"explanation":"Probability measures how likely an outcome or event is; it is not the result itself. One possible result of an experiment is an outcome."}]},{"id":"2eb01134-4dcb-da13-89c5-143c2c4265ec","cardType":"basic","points":1,"front":"A cafe offers 2 sandwiches and 3 drinks. How many lunch combinations are possible?","back":"6 (calculated as $2 \\times 3$)","mcqOptions":[{"text":"6 (calculated as $2 \\times 3$)","correct":true,"explanation":"Each of the 2 sandwiches can be paired with any of the 3 drinks, so the multiplication principle gives $2 \\times 3 = 6$ combinations."},{"text":"5 (calculated as $2 + 3$)","correct":false,"explanation":"Adding counts the five individual menu items, not the sandwich-and-drink pairings. Each sandwich pairs with three drinks, so $2 \\times 3 = 6$ combinations."},{"text":"3 (calculated as $2 + 3 - 2$)","correct":false,"explanation":"This subtraction incorrectly removes possible pairings and effectively considers only one sandwich. Both sandwiches pair with all three drinks, giving $2 \\times 3 = 6$ combinations."},{"text":"9 (calculated as $3 \\times 3$)","correct":false,"explanation":"This mistakenly uses three sandwich choices as well as three drinks. There are only two sandwiches, so the correct calculation is $2 \\times 3 = 6$."}]},{"id":"5be0764a-7cc7-f17d-c1ff-17b57202757e","cardType":"cloze","points":null,"front":"To keep a list neat for a coin toss, you can use the short labels [ ] and T.","back":"To keep a list neat for a coin toss, you can use the short labels H and T.","mcqOptions":[{"text":"H","correct":true,"explanation":"H is the standard short label for heads. Together, H and T allow coin-toss sequences to be recorded clearly and compactly."},{"text":"C","correct":false,"explanation":"C might be chosen to represent coin, but it does not identify which side appears. The standard label for heads is H, paired with T for tails."},{"text":"A","correct":false,"explanation":"A is not a standard label for either result of a coin toss. Heads is abbreviated to H, while tails is abbreviated to T."},{"text":"X","correct":false,"explanation":"X could mark an unspecified result, but it does not communicate that heads occurred. The conventional short label for heads is H."}]},{"id":"0c9495b4-802b-f8b2-95fa-704006342fdf","cardType":"cloze","points":null,"front":"To keep a list neat for a coin toss, you can use the short labels H and [ ].","back":"To keep a list neat for a coin toss, you can use the short labels H and T.","mcqOptions":[{"text":"T","correct":true,"explanation":"T is the standard short label for tails. Using H and T makes it easy to write and systematically list sequences of coin-toss outcomes."},{"text":"C","correct":false,"explanation":"C may suggest coin, but it does not distinguish tails from heads. The conventional short label for tails is T, used alongside H."},{"text":"B","correct":false,"explanation":"B is not the standard abbreviation for a coin-toss result. Tails is represented by T, while heads is represented by H."},{"text":"X","correct":false,"explanation":"X usually represents an unknown or unspecified value rather than tails. The standard short label used with H is T."}]},{"id":"b1aa17c9-c43f-1c9c-59f2-4b6288cd2f03","cardType":"basic","points":2,"front":"Why are HHT and HTH considered different outcomes when tossing a coin three times?","back":"Because order matters: the tails happens on a different toss.","mcqOptions":[{"text":"Because order matters: the tails happens on a different toss.","correct":true,"explanation":"HHT has tails on the third toss, whereas HTH has tails on the second toss. Since the position of each result matters, the sequences are distinct outcomes."},{"text":"Because the totals differ: one sequence contains more heads than the other.","correct":false,"explanation":"Both sequences contain two heads and one tail, so their totals are identical. They are different because order matters and the tail occurs on a different toss."},{"text":"Because their probabilities differ: HHT is more likely to occur than HTH.","correct":false,"explanation":"For a fair coin, both specified sequences have probability $(\\frac{1}{2})^3 = \\frac{1}{8}$. They differ only because their heads and tails occur in different orders."},{"text":"Because tails becomes less likely: its probability changes on each successive toss.","correct":false,"explanation":"Independent coin tosses do not become more or less likely based on their position. HHT and HTH are distinct because the tail occurs on different tosses."}]},{"id":"9adcfbaa-0014-828b-014c-3691e4d25838","cardType":"basic","points":1,"front":"How many total outcomes are in the sample space for tossing a coin three times?","back":"8","mcqOptions":[{"text":"8","correct":true,"explanation":"Each of the three tosses has 2 possible results, H or T. The multiplication principle therefore gives $2 \\times 2 \\times 2 = 8$ outcomes."},{"text":"6","correct":false,"explanation":"Multiplying the 2 possible results by the 3 tosses gives 6, but choices across successive tosses multiply. The correct calculation is $2^3 = 8$."},{"text":"9","correct":false,"explanation":"Squaring the number of tosses gives $3^2 = 9$, but there are two choices on each of three tosses. Therefore, the total is $2^3 = 8$."},{"text":"4","correct":false,"explanation":"This counts only two tosses, since $2^2 = 4$. Including the third independent toss doubles the number of sequences, giving $2^3 = 8$."}]},{"id":"adde9647-a978-2470-8455-f733d9c1b9a6","cardType":"cloze","points":null,"front":"When making numbers from cards, each card can usually be used [ ] unless the question says otherwise.","back":"When making numbers from cards, each card can usually be used once unless the question says otherwise.","mcqOptions":[{"text":"once","correct":true,"explanation":"Each physical card represents one available digit, so it can normally be selected only once when constructing a number unless repetition is explicitly permitted."},{"text":"twice","correct":false,"explanation":"Using every card twice incorrectly treats each card as reusable. The correct term is “once” because each physical card can normally be used only one time."},{"text":"repeatedly","correct":false,"explanation":"Unlimited repetition ignores the fixed supply of number cards. The blank should be filled by “once”, as a card is unavailable after it has been used."},{"text":"more than once","correct":false,"explanation":"This incorrectly allows a card to be selected again. Each card can usually be used “once”, although the available cards may be arranged in any order."}]},{"id":"e1f50ad3-4f82-3b4c-82be-d6ce034adb8f","cardType":"basic","points":1,"front":"If given number cards 4, 6, and 9, why is the number \"44\" usually not allowed?","back":"There is only one card with the digit 4.","mcqOptions":[{"text":"There is only one card with the digit 4.","correct":true,"explanation":"Making 44 requires two separate cards showing 4. Because the set contains only one such card, the digit 4 cannot be selected twice."},{"text":"There are two cards with the digit 4.","correct":false,"explanation":"Having two cards showing 4 would make 44 possible, not prevent it. The actual restriction is that there is only one card with the digit 4."},{"text":"The cards must be used in ascending order.","correct":false,"explanation":"Number cards do not normally have to be arranged in ascending order. The number 44 is disallowed because forming it requires two cards showing 4, but only one is available."},{"text":"Only odd 2-digit numbers are allowed.","correct":false,"explanation":"There is no general rule restricting the result to odd numbers. The issue is that 44 needs two separate 4 cards, while the set contains only one."}]},{"id":"783ace62-eec9-66f1-e2b7-ab92442021f7","cardType":"basic","points":2,"front":"List all different 3-digit numbers that can be made using cards 2, 2, and 5.","back":"

","mcqOptions":[{"text":"

","correct":true,"explanation":"The three distinct arrangements are obtained by placing the 5 in the final, middle and first positions. Swapping the identical 2s creates no new number."},{"text":"

","correct":false,"explanation":"The number 525 requires two cards showing 5, but only one is available. The valid unique arrangements are 225, 252 and 522."},{"text":"

","correct":false,"explanation":"The number 552 incorrectly uses two 5 cards and only one 2 card. Using exactly the supplied cards gives 225, 252 and 522."},{"text":"

","correct":false,"explanation":"The number 525 does not match the available cards because it contains two 5s. The complete valid list is 225, 252 and 522."}]},{"id":"04d5a96c-6e2a-d7cc-b3c3-68391c2b484f","cardType":"cloze","points":null,"front":"When listing matches between teams, [ ] because Lions v Tigers is the same match as Tigers v Lions.","back":"When listing matches between teams, order does not matter because Lions v Tigers is the same match as Tigers v Lions.","mcqOptions":[{"text":"order does not matter","correct":true,"explanation":"A match is determined by the pair of teams involved, not by which team is named first. Reversing the names therefore does not create another match."},{"text":"order matters","correct":false,"explanation":"Treating order as important would count Lions v Tigers and Tigers v Lions separately. The correct fill is “order does not matter” because both name the same pairing."},{"text":"teams can play twice","correct":false,"explanation":"The statement concerns whether reversed team names represent a new pairing, not how often teams play. The correct fill is “order does not matter”."},{"text":"each ordering is counted separately","correct":false,"explanation":"Counting each ordering separately would double-count every match. The correct fill is “order does not matter” because reversing the two teams leaves the pairing unchanged."}]},{"id":"13933980-7fc3-7ab3-7536-c1499b740350","cardType":"basic","points":3,"front":"List all unique matches if four teams (A, B, C, D) play each other once.","back":"

A v B, A v C, A v D
B v C, B v D
C v D

","mcqOptions":[{"text":"

A v B, A v C, A v D
B v C, B v D
C v D

","correct":true,"explanation":"Each unordered pair appears exactly once, producing six matches altogether. Reversing a pairing would name the same match and must not be counted again."},{"text":"

A v B, A v C, A v D
B v A, B v C
C v D

","correct":false,"explanation":"B v A duplicates A v B, while B v D is missing. Reversed team order does not create a new match, so the correct list contains each of the six pairs once."},{"text":"

A v B, A v C, A v D
B v C, B v D
A v A

","correct":false,"explanation":"A v A is not a match between two different teams, and C v D has been omitted. Four teams produce six distinct pairings with no self-matches."},{"text":"

A v B, A v C, A v D
B v C, B v D
C v B

","correct":false,"explanation":"C v B repeats B v C and leaves out C v D. Each pair must appear once regardless of ordering, giving six unique matches in total."}]},{"id":"0bd9340e-9727-4977-144b-b08d90d18bd2","cardType":"basic","points":1,"front":"What does it mean for outcomes to be equally likely?","back":"Each outcome has the same chance of happening.","mcqOptions":[{"text":"Each outcome has the same chance of happening.","correct":true,"explanation":"Equally likely outcomes have identical probabilities, even though a short sequence of trials may not produce each outcome the same number of times."},{"text":"Each outcome happens the same number of times in every experiment.","correct":false,"explanation":"Equal frequencies in every experiment are not guaranteed because random variation occurs. Equally likely means that each outcome has the same chance before the experiment."},{"text":"Each outcome has a different chance of happening.","correct":false,"explanation":"Different probabilities are the opposite of being equally likely. The correct definition is that every outcome has the same chance of happening."},{"text":"At least one outcome has a 50% chance of happening.","correct":false,"explanation":"A 50% outcome does not make all outcomes equally likely, especially when there are more than two possibilities. Every outcome must instead have the same chance."}]},{"id":"d9ccb203-fc57-dcff-c568-1187c46b9fa4","cardType":"basic","points":1,"front":"What is the formula for probability using a sample space of equally likely outcomes?","back":"$$$\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}$$","mcqOptions":[{"text":"$$$\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}$$","correct":true,"explanation":"For equally likely outcomes, probability is the proportion of all possible outcomes that satisfy the event, so favourable outcomes form the numerator and total outcomes the denominator."},{"text":"$$$\\frac{\\text{number of favourable outcomes}}{\\text{number of unfavourable outcomes}}$$","correct":false,"explanation":"Dividing by unfavourable outcomes calculates favourable-to-unfavourable odds, not probability. Probability divides the number of favourable outcomes by the total number of outcomes."},{"text":"$$$\\frac{\\text{total number of outcomes}}{\\text{number of favourable outcomes}}$$","correct":false,"explanation":"This reverses the required fraction and can produce values greater than 1. The favourable outcomes must be divided by the total number of outcomes."},{"text":"$$$\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes} - \\text{number of favourable outcomes}}$$","correct":false,"explanation":"Subtracting favourable outcomes leaves only the unfavourable outcomes in the denominator, producing odds rather than probability. The denominator should be the total number of outcomes."}]},{"id":"36d765d6-b5c3-934d-cd4c-437a5b37bda3","cardType":"basic","points":1,"front":"What is the neatest way to organise the sample space for two dice?","back":"A grid (with one dice across the top and one down the side).","mcqOptions":[{"text":"A grid (with one dice across the top and one down the side).","correct":true,"explanation":"A grid pairs every result on the first die with every result on the second die, displaying all 36 ordered outcomes systematically without omissions."},{"text":"A tree diagram (with one dice on each separate branch).","correct":false,"explanation":"A tree diagram can represent repeated events, but it is less neat for the 36 pairs from two dice. A grid displays every pairing directly."},{"text":"A tally chart (with each dice total listed in one column).","correct":false,"explanation":"A tally chart records observed frequencies rather than systematically listing the theoretical sample space. A grid with one die on each axis shows all possible pairs."},{"text":"A Venn diagram (with one dice represented by each circle).","correct":false,"explanation":"A Venn diagram represents relationships between sets, not every ordered pair of dice results. A grid is the clearest way to organise the 36 outcomes."}]},{"id":"ba0068f7-b37a-113e-81e8-cef0d299aed3","cardType":"cloze","points":null,"front":"A [ ] is an outcome that matches what the question wants.","back":"A favourable outcome is an outcome that matches what the question wants.","mcqOptions":[{"text":"favourable outcome","correct":true,"explanation":"A favourable outcome satisfies the condition being investigated. Probability is often calculated by dividing the number of favourable outcomes by the total number of possible outcomes."},{"text":"possible outcome","correct":false,"explanation":"A possible outcome is any result that could occur, whether or not it matches the question’s condition. An outcome that does match the condition is a favourable outcome."},{"text":"equally likely outcome","correct":false,"explanation":"Equally likely outcomes have the same probability of occurring, but they need not match the condition in the question. A matching result is called a favourable outcome."},{"text":"mutually exclusive outcome","correct":false,"explanation":"Mutually exclusive outcomes cannot occur together in the same trial. The term for an outcome that matches the condition being investigated is a favourable outcome."}]},{"id":"85f206d9-6c0e-c6ae-9174-c7c8e60a39c5","cardType":"basic","points":2,"front":"Box A has cards 1, 4, 7. Box B has 3, 5, 6. If one is picked from each, when is the product odd?","back":"When both cards picked are odd numbers.","mcqOptions":[{"text":"When both cards picked are odd numbers.","correct":true,"explanation":"A product is odd only when every factor is odd. Therefore, the card selected from Box A and the card selected from Box B must both be odd."},{"text":"When both cards picked are even numbers.","correct":false,"explanation":"Multiplying two even numbers produces an even product, not an odd one. The product is odd only when both selected cards are odd numbers."},{"text":"When at least one card picked is an odd number.","correct":false,"explanation":"Having one odd card is insufficient because an odd number multiplied by an even number is even. Both selected cards must be odd for the product to be odd."},{"text":"When one card picked is odd and the other is even.","correct":false,"explanation":"An odd number multiplied by an even number always gives an even product. To obtain an odd product, both cards picked must be odd numbers."}]},{"id":"0d2321ca-fda9-5a49-3dc5-e17ab0e25bc4","cardType":"basic","points":2,"front":"List the two-digit numbers possible from cards 4, 6, and 9.","back":"

46, 49
64, 69
94, 96

","mcqOptions":[{"text":"

46, 49
64, 69
94, 96

","correct":true,"explanation":"Each card is used once in each number, and each of the three digits can occupy the tens place. This systematically produces all six possible two-digit numbers."},{"text":"

46, 49
64, 69
94, 99

","correct":false,"explanation":"The number 99 incorrectly uses the 9 card twice, while 96 has been omitted. Using two different cards in both possible orders gives 46, 49, 64, 69, 94 and 96."},{"text":"

44, 46
49, 64
69, 96

","correct":false,"explanation":"The number 44 incorrectly repeats the 4 card, and 94 is missing. The complete list is 46, 49, 64, 69, 94 and 96."},{"text":"

46, 49
64, 66
94, 96

","correct":false,"explanation":"The number 66 incorrectly uses the 6 card twice, while 69 has been missed. Systematic listing gives 46, 49, 64, 69, 94 and 96."}]},{"id":"80fee421-0d1f-c00c-1c74-ffd3e60fe31a","cardType":"cloze","points":null,"front":"[ ] means writing the sample space in an organised order to avoid missing or repeating answers.","back":"Systematic listing means writing the sample space in an organised order to avoid missing or repeating answers.","mcqOptions":[{"text":"Systematic listing","correct":true,"explanation":"Systematic listing arranges possible outcomes in a logical order. This helps ensure that every outcome is included exactly once when constructing the sample space."},{"text":"Systematic sampling","correct":false,"explanation":"Systematic sampling selects members of a population at regular intervals, such as every tenth person. Writing all possible outcomes in an organised order is systematic listing."},{"text":"Tree diagram","correct":false,"explanation":"A tree diagram represents successive choices using branches, rather than defining the process of writing outcomes in an organised list. That process is systematic listing."},{"text":"Two-way table","correct":false,"explanation":"A two-way table organises information using rows and columns for two categories. The specific method of writing the sample space in an ordered list is systematic listing."}]},{"id":"39146379-ff44-0442-97ac-de1de00b61e0","cardType":"basic","points":3,"front":"In a systematic list of 3 coin tosses, how do you find the probability of getting exactly 2 heads?","back":"

Identify all 8 outcomes
Count outcomes with two H (HHT, HTH, THH)
Probability is $\\frac{3}{8}$

","mcqOptions":[{"text":"

Identify all 8 outcomes
Count outcomes with two H (HHT, HTH, THH)
Probability is $\\frac{3}{8}$

","correct":true,"explanation":"Three tosses have $2^3=8$ equally likely outcomes. Exactly three of these, HHT, HTH and THH, contain two heads, so the probability is $\\frac{3}{8}$."},{"text":"

Identify all 8 outcomes
Count outcomes with at least two H (HHT, HTH, THH, HHH)
Probability is $\\frac{4}{8}$

","correct":false,"explanation":"This includes HHH by changing “exactly two” to “at least two”. Only HHT, HTH and THH are favourable, so the probability is $\\frac{3}{8}$."},{"text":"

Identify all 8 outcomes
Count outcomes with consecutive H (HHT, THH)
Probability is $\\frac{2}{8}$

","correct":false,"explanation":"This wrongly requires the two heads to be consecutive, excluding HTH. All three arrangements HHT, HTH and THH count, giving a probability of $\\frac{3}{8}$."},{"text":"

Identify the 4 possible head totals
Select the total with two H
Probability is $\\frac{1}{4}$

","correct":false,"explanation":"The four possible head totals are not equally likely, so they cannot be used as equally weighted outcomes. Listing all eight sequences gives three favourable outcomes and probability $\\frac{3}{8}$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Systematic Listing","title":"Systematic Listing Flashcards","topbar":{"backHref":"/gcse/maths/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"b192022f-4321-4322-9bba-c31cb494bd58","walled":"$L38"}],"$L39","$L3a"]