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Probability 0 represents an impossible event, while probability 1 represents a certain event."},{"text":"From -1 to 1 inclusive.","correct":false,"explanation":"Including negative values confuses probability with scales that permit signed numbers. Every probability lies from 0 to 1 inclusive, because chances cannot be below impossibility."},{"text":"From 0 to 0.5 inclusive.","correct":false,"explanation":"This excludes probabilities greater than one half, including likely and certain events. The correct interval is from 0 to 1 inclusive."},{"text":"From 0 to 100 inclusive.","correct":false,"explanation":"Probabilities are not percentages unless converted, so the upper bound is 1 rather than 100. The correct interval is from 0 to 1 inclusive."}]},{"id":"a464ea07-eaae-40f0-8000-662e0288e628","cardType":"cloze","points":null,"front":"Rolling a 5 is an [ ]; rolling an even number is an [ ].","back":"Rolling a 5 is an outcome; rolling an even number is an event.","mcqOptions":[{"text":"Rolling a 5 is an outcome; rolling an even number is an event.","correct":true,"explanation":"An outcome is one specific possible result, such as rolling 5. An event is a set of outcomes, such as rolling any even number."},{"text":"Rolling a 5 is an event; rolling an even number is an outcome.","correct":false,"explanation":"This reverses the definitions. Rolling 5 is one specific outcome, whereas rolling an even number describes the event containing outcomes 2, 4 and 6."},{"text":"Rolling a 5 is a trial; rolling an even number is an outcome.","correct":false,"explanation":"A trial is the action of rolling the die, not the result obtained. Rolling 5 is an outcome, while rolling an even number is an event."},{"text":"Rolling a 5 is a sample space; rolling an even number is a sample space.","correct":false,"explanation":"A sample space contains all possible results, so neither description is a sample space. Rolling 5 is an outcome, and rolling an even number is an event."}]},{"id":"f2e23582-0309-4330-8fe4-530c9767111b","cardType":"cloze","points":null,"front":"A result that is not chosen on purpose is [ ].","back":"A result that is not chosen on purpose is at random.","mcqOptions":[{"text":"at random","correct":true,"explanation":"A result obtained without deliberately selecting it is described as being at random. Random selection is intended to avoid purposeful choice influencing the result."},{"text":"biased","correct":false,"explanation":"A biased result is affected by an unfair method or unequal chances, not merely unchosen. A result not chosen on purpose is described as being at random."},{"text":"systematic","correct":false,"explanation":"A systematic result follows a planned rule or pattern, which conflicts with selection without purposeful choice. The required description is at random."},{"text":"certain","correct":false,"explanation":"A certain result must occur, whereas lack of purposeful selection does not guarantee any particular result. The correct description is at random."}]},{"id":"9c36a1b3-dfbb-4f18-8c19-e360a430de9b","cardType":"cloze","points":null,"front":"If each outcome has the same chance, the outcomes are [ ].","back":"If each outcome has the same chance, the outcomes are equally likely.","mcqOptions":[{"text":"equally likely","correct":true,"explanation":"Outcomes are equally likely when every one has the same probability of occurring. For example, each face of a fair six-sided die has probability $\\frac{1}{6}$."},{"text":"unequally likely","correct":false,"explanation":"Unequally likely means that some outcomes have different probabilities, directly contradicting the condition that each has the same chance. The outcomes are equally likely."},{"text":"certain","correct":false,"explanation":"Certain means having probability 1, rather than sharing the same chance with other possible outcomes. Outcomes with the same probability are equally likely."},{"text":"impossible","correct":false,"explanation":"Impossible outcomes each have probability 0 and cannot occur; equal non-zero chances are not described this way. Outcomes with the same chance are equally likely."}]},{"id":"5198de41-65a4-4ee2-a855-415c4f92eaf2","cardType":"basic","points":1,"front":"For equally likely outcomes, what calculation gives the probability of an event?","back":"$\\frac{\\text{wanted outcomes}}{\\text{total outcomes}}$","mcqOptions":[{"text":"$\\frac{\\text{wanted outcomes}}{\\text{total outcomes}}$","correct":true,"explanation":"For equally likely outcomes, probability is found by dividing the number of outcomes that satisfy the event by the total number of possible outcomes."},{"text":"$\\frac{\\text{total outcomes}}{\\text{wanted outcomes}}$","correct":false,"explanation":"Reversing the fraction can produce a value greater than 1, which cannot be a probability. Divide wanted outcomes by total outcomes instead."},{"text":"$\\frac{\\text{unwanted outcomes}}{\\text{total outcomes}}$","correct":false,"explanation":"This calculates the probability of the complementary event by counting outcomes that are not wanted. The required probability uses wanted outcomes over total outcomes."},{"text":"$\\frac{\\text{wanted outcomes}}{\\text{unwanted outcomes}}$","correct":false,"explanation":"Comparing wanted outcomes only with unwanted outcomes omits part of the sample space from the denominator. The denominator must be the total number of outcomes."}]},{"id":"7fb3c3a6-c332-4351-8903-f283e005ceb1","cardType":"basic","points":1,"front":"Cards show 2, 4, 4, 5, 7, 8, 9. What is the probability of picking a 4?","back":"$\\frac{2}{7}$","mcqOptions":[{"text":"$\\frac{2}{7}$","correct":true,"explanation":"There are seven cards in total and two of them show 4. Therefore, the probability of selecting a 4 is $\\frac{2}{7}$."},{"text":"$\\frac{1}{7}$","correct":false,"explanation":"This counts the value 4 only once because it treats repeated values as a single outcome. Two of the seven cards show 4, giving $\\frac{2}{7}$."},{"text":"$\\frac{2}{6}$","correct":false,"explanation":"This incorrectly removes one of the two matching cards from the total after counting it. All seven cards remain possible selections, so the probability is $\\frac{2}{7}$."},{"text":"$\\frac{4}{7}$","correct":false,"explanation":"This uses the number printed on the desired cards as the numerator instead of counting those cards. There are two matching cards out of seven, giving $\\frac{2}{7}$."}]},{"id":"99999f11-d7de-4794-82fd-5df8ec01ff45","cardType":"cloze","points":null,"front":"On a probability scale, 0 means [ ] and 1 means [ ].","back":"On a probability scale, 0 means impossible and 1 means certain.","mcqOptions":[{"text":"On a probability scale, 0 means impossible and 1 means certain.","correct":true,"explanation":"Probability 0 represents an event that cannot happen, so it is impossible. Probability 1 represents an event that must happen, so it is certain."},{"text":"On a probability scale, 0 means certain and 1 means impossible.","correct":false,"explanation":"This reverses the endpoints of the probability scale. Probability 0 means impossible because the event cannot occur, while probability 1 means certain."},{"text":"On a probability scale, 0 means unlikely and 1 means likely.","correct":false,"explanation":"Unlikely and likely describe probabilities near the endpoints, not exactly at them. Probability 0 means impossible, and probability 1 means certain."},{"text":"On a probability scale, 0 means an even chance and 1 means an even chance.","correct":false,"explanation":"An even chance corresponds to probability $\\frac{1}{2}$, so it cannot describe either endpoint. At 0 an event is impossible, and at 1 it is certain."}]},{"id":"bb2acd35-7a44-4bed-8652-ef440232876d","cardType":"cloze","points":null,"front":"On a probability scale, [ ] means an even chance.","back":"On a probability scale, $\\frac{1}{2}$ means an even chance.","mcqOptions":[{"text":"$$\\frac{1}{2}$","correct":true,"explanation":"An even chance means that an event is equally likely to happen or not happen. Its probability is therefore one half, written $\\frac{1}{2}$."},{"text":"$$0$","correct":false,"explanation":"A probability of 0 represents an impossible event, not an even chance. An event with an even chance has probability $\\frac{1}{2}$."},{"text":"$$1$","correct":false,"explanation":"A probability of 1 represents a certain event, not one equally likely to happen or not happen. An even chance is represented by $\\frac{1}{2}$."},{"text":"$$\\frac{1}{4}$","correct":false,"explanation":"This gives a one-in-four chance, so the event is less likely to happen than not happen. An even chance has probability $\\frac{1}{2}$."}]},{"id":"34974dd6-27f8-4f5c-9461-934fdae0f36c","cardType":"basic","points":1,"front":"On a probability scale, where is $\\frac{1}{4}$ placed?","back":"Halfway between 0 and $\\frac{1}{2}$.","mcqOptions":[{"text":"Halfway between 0 and $\\frac{1}{2}$.","correct":true,"explanation":"The midpoint of 0 and $\\frac{1}{2}$ is $\\frac{1}{4}$, because half of $\\frac{1}{2}$ is $\\frac{1}{4}$."},{"text":"Halfway between 0 and 1.","correct":false,"explanation":"The midpoint of 0 and 1 is $\\frac{1}{2}$, not $\\frac{1}{4}$. The value $\\frac{1}{4}$ is halfway between 0 and $\\frac{1}{2}$."},{"text":"Halfway between $\\frac{1}{4}$ and $\\frac{1}{2}$.","correct":false,"explanation":"The midpoint of $\\frac{1}{4}$ and $\\frac{1}{2}$ is $\\frac{3}{8}$. The value $\\frac{1}{4}$ is halfway between 0 and $\\frac{1}{2}$."},{"text":"Halfway between 0 and $\\frac{1}{4}$.","correct":false,"explanation":"The midpoint of 0 and $\\frac{1}{4}$ is $\\frac{1}{8}$, which is too small. The value $\\frac{1}{4}$ lies halfway between 0 and $\\frac{1}{2}$."}]},{"id":"ccbf5f9f-f6db-4732-b2f8-22d067f58464","cardType":"basic","points":1,"front":"On a probability scale, where is $\\frac{3}{4}$ placed?","back":"Halfway between $\\frac{1}{2}$ and 1.","mcqOptions":[{"text":"Halfway between $\\frac{1}{2}$ and 1.","correct":true,"explanation":"The gap from $\\frac{1}{2}$ to 1 is $\\frac{1}{2}$, and moving halfway across it gives $\\frac{1}{2}+\\frac{1}{4}=\\frac{3}{4}$."},{"text":"Halfway between 0 and 1.","correct":false,"explanation":"The midpoint of the full probability scale from 0 to 1 is $\\frac{1}{2}$. The value $\\frac{3}{4}$ is halfway between $\\frac{1}{2}$ and 1."},{"text":"Halfway between 0 and $\\frac{3}{4}$.","correct":false,"explanation":"The midpoint of 0 and $\\frac{3}{4}$ is $\\frac{3}{8}$, not $\\frac{3}{4}$. The correct endpoints are $\\frac{1}{2}$ and 1."},{"text":"Halfway between $\\frac{3}{4}$ and 1.","correct":false,"explanation":"The midpoint of $\\frac{3}{4}$ and 1 is $\\frac{7}{8}$, so this places the point too high. $\\frac{3}{4}$ is halfway between $\\frac{1}{2}$ and 1."}]},{"id":"4a72e5d0-b21e-4cad-a707-21c243e06c36","cardType":"basic","points":1,"front":"A fair dice is rolled once. What is the probability of rolling an 8?","back":"0","mcqOptions":[{"text":"0","correct":true,"explanation":"A standard fair dice has outcomes 1 to 6, so rolling an 8 is impossible. An impossible event has probability 0."},{"text":"$\\frac{1}{6}$","correct":false,"explanation":"The probability $\\frac{1}{6}$ applies to one possible face, such as rolling a 4. Since 8 is not on a standard dice, its probability is 0."},{"text":"$\\frac{1}{8}$","correct":false,"explanation":"Using 8 as the denominator mistakes the requested number for the number of possible outcomes. A dice has six outcomes and none is 8, so the probability is 0."},{"text":"1","correct":false,"explanation":"A probability of 1 describes a certain event, but rolling an 8 on a standard six-sided dice is impossible. Its probability is therefore 0."}]},{"id":"3369a649-e006-4830-afca-04871893f3e3","cardType":"basic","points":1,"front":"A fair dice is rolled once. What is the probability of rolling a number less than 7?","back":"1","mcqOptions":[{"text":"1","correct":true,"explanation":"Every possible result on a standard dice - 1, 2, 3, 4, 5 or 6 - is less than 7, so the event is certain and has probability 1."},{"text":"$\\frac{5}{6}$","correct":false,"explanation":"This commonly comes from leaving out 6, but 6 is less than 7 and must be included. All six outcomes qualify, giving $\\frac{6}{6}=1$."},{"text":"$\\frac{6}{7}$","correct":false,"explanation":"Using 7 as the denominator confuses the comparison value with the number of dice outcomes. All six possible outcomes are less than 7, so the probability is 1."},{"text":"$\\frac{1}{6}$","correct":false,"explanation":"This counts only one successful outcome, but all six faces are numbered below 7. The event is certain, so its probability is $\\frac{6}{6}=1$."}]},{"id":"98e8f879-931d-41e0-933a-2b294769fd3c","cardType":"basic","points":1,"front":"A fair dice is rolled once. What is the probability of rolling a number greater than 3?","back":"$\\frac{1}{2}$","mcqOptions":[{"text":"$\\frac{1}{2}$","correct":true,"explanation":"The results greater than 3 are 4, 5 and 6. That gives three favourable outcomes out of six, so the probability is $\\frac{3}{6}=\\frac{1}{2}$."},{"text":"$\\frac{1}{3}$","correct":false,"explanation":"This counts only two successful outcomes, often by overlooking 4. The results greater than 3 are 4, 5 and 6, giving $\\frac{3}{6}=\\frac{1}{2}$."},{"text":"$\\frac{2}{3}$","correct":false,"explanation":"This usually comes from including 3 among the successful outcomes. Greater than 3 excludes 3, leaving 4, 5 and 6, so the probability is $\\frac{1}{2}$."},{"text":"$\\frac{3}{5}$","correct":false,"explanation":"The denominator must include all six possible dice outcomes, not only five of them. Three outcomes are greater than 3, so the probability is $\\frac{3}{6}=\\frac{1}{2}$."}]},{"id":"663b2936-d5af-4740-b010-d80b4babcbf9","cardType":"cloze","points":null,"front":"In probability, less than 3 means 1 and 2, so it [ ].","back":"In probability, less than 3 means 1 and 2, so it excludes 3.","mcqOptions":[{"text":"excludes 3","correct":true,"explanation":"The phrase “less than 3” includes only values strictly below 3. For whole-number dice outcomes, these are 1 and 2, so 3 itself is excluded."},{"text":"includes 3","correct":false,"explanation":"Including 3 confuses “less than” with “less than or equal to”. A strict inequality does not include its boundary value, so less than 3 excludes 3."},{"text":"excludes 2","correct":false,"explanation":"The number 2 is below 3, so it must be included. Less than 3 includes 1 and 2 but excludes the boundary value 3."},{"text":"includes only 3","correct":false,"explanation":"This confuses values below 3 with the value equal to 3. Less than 3 includes 1 and 2 and excludes 3 itself."}]},{"id":"9bbc9029-89a3-4aea-ba13-6289e9f4fade","cardType":"basic","points":1,"front":"A box has 25 counters: 9 red, 6 blue and the rest green. How many are green?","back":"10","mcqOptions":[{"text":"10","correct":true,"explanation":"The red and blue counters total $9+6=15$. Subtracting these from the 25 counters leaves $25-15=10$ green counters."},{"text":"9","correct":false,"explanation":"This repeats the number of red counters instead of finding the remainder. After accounting for 9 red and 6 blue counters, $25-9-6=10$ are green."},{"text":"15","correct":false,"explanation":"This is the combined number of red and blue counters, not the number of green counters. Subtracting it from 25 gives $25-15=10$ green counters."},{"text":"16","correct":false,"explanation":"This comes from subtracting only the 9 red counters and forgetting the 6 blue counters. Both known colours must be removed, giving $25-9-6=10$."}]},{"id":"6f49d0cc-3fc0-463f-a1bb-23b5b26d83e7","cardType":"basic","points":2,"front":"A box has 25 counters: 9 red, 6 blue and the rest green. What is the probability of green?","back":"$\\frac{2}{5}$","mcqOptions":[{"text":"$\\frac{2}{5}$","correct":true,"explanation":"There are $25-9-6=10$ green counters, so the probability is $\\frac{10}{25}$. Dividing numerator and denominator by 5 simplifies this to $\\frac{2}{5}$."},{"text":"$\\frac{3}{5}$","correct":false,"explanation":"This is the probability of selecting a counter that is not green, since there are 15 red or blue counters. Green has probability $\\frac{10}{25}=\\frac{2}{5}$."},{"text":"$\\frac{2}{3}$","correct":false,"explanation":"This incorrectly uses the 15 red and blue counters as the total. The denominator must be all 25 counters, so green has probability $\\frac{10}{25}=\\frac{2}{5}$."},{"text":"$\\frac{2}{25}$","correct":false,"explanation":"This simplifies the numerator from 10 to 2 without also simplifying the denominator. Dividing both parts of $\\frac{10}{25}$ by 5 gives $\\frac{2}{5}$."}]},{"id":"fc420b1e-fb5e-4709-ac1b-af6bb0d301e4","cardType":"basic","points":1,"front":"A bag has 5 red counters and 1 blue counter. What is the probability of red?","back":"$\\frac{5}{6}$","mcqOptions":[{"text":"$\\frac{5}{6}$","correct":true,"explanation":"There are 6 counters altogether, and 5 of them are red. Therefore, the probability of selecting red is $\\frac{5}{6}$."},{"text":"$\\frac{1}{6}$","correct":false,"explanation":"This is the probability of selecting the single blue counter. Red has 5 favourable outcomes out of 6 counters, so its probability is $\\frac{5}{6}$."},{"text":"$\\frac{5}{5}$","correct":false,"explanation":"This uses only the five red counters as the denominator, incorrectly suggesting that every counter is red. The denominator must be the total of 6, giving $\\frac{5}{6}$."},{"text":"$\\frac{6}{5}$","correct":false,"explanation":"This reverses the required fraction, placing the total number above the number of favourable outcomes. Probability is favourable outcomes over total outcomes, so it is $\\frac{5}{6}$."}]},{"id":"ff36b276-ed3a-4368-a06f-18b0e1c4a2f4","cardType":"cloze","points":null,"front":"An event and its complement have probabilities that add to [ ].","back":"An event and its complement have probabilities that add to 1.","mcqOptions":[{"text":"1","correct":true,"explanation":"An event and its complement cover all possible outcomes without overlapping. Their probabilities must therefore add to 1, representing certainty."},{"text":"0","correct":false,"explanation":"A total probability of 0 represents an impossible event, not all possible outcomes. An event and its complement together are certain, so their probabilities add to 1."},{"text":"2","correct":false,"explanation":"Probabilities across all mutually exclusive possible outcomes cannot total 2. An event and its complement exhaust the sample space, so their probabilities add to 1."},{"text":"0.5","correct":false,"explanation":"A total of 0.5 would represent only half of the total probability. An event and its complement include every outcome, so their probabilities add to 1."}]},{"id":"e98a6c49-faa0-422f-b75f-4ce663da23d5","cardType":"basic","points":1,"front":"A player has probability 0.7 of winning. What is the probability of not winning?","back":"0.3","mcqOptions":[{"text":"0.3","correct":true,"explanation":"Winning and not winning are complementary events, so their probabilities add to 1. Therefore, the probability of not winning is $1 - 0.7 = 0.3$."},{"text":"0.7","correct":false,"explanation":"This repeats the probability of winning rather than finding its complement. Subtracting the winning probability from 1 gives $1 - 0.7 = 0.3$."},{"text":"1.7","correct":false,"explanation":"This adds the winning probability to 1, producing a value greater than the maximum possible probability. The complement is found by $1 - 0.7 = 0.3$."},{"text":"0.03","correct":false,"explanation":"This introduces an unnecessary decimal-place shift after finding the complement. The correct subtraction is $1.0 - 0.7 = 0.3$, not 0.03."}]},{"id":"33b13ad6-8c2a-4dbd-b360-7882bf8cd6ec","cardType":"basic","points":2,"front":"A table shows 6 red, 4 blue, 5 yellow and 3 green counters. What is the probability of not blue?","back":"$\\frac{7}{9}$","mcqOptions":[{"text":"$\\frac{7}{9}$","correct":true,"explanation":"There are 18 counters in total and 14 are not blue. Thus the probability is $\\frac{14}{18}$, which simplifies to $\\frac{7}{9}$."},{"text":"$\\frac{2}{9}$","correct":false,"explanation":"This is the probability of selecting blue, since 4 of the 18 counters are blue. The complement is $1 - \\frac{2}{9} = \\frac{7}{9}$."},{"text":"$\\frac{7}{18}$","correct":false,"explanation":"This correctly identifies 14 non-blue counters but incorrectly uses 36 rather than 18 as the total before simplifying. The correct probability is $\\frac{14}{18} = \\frac{7}{9}$."},{"text":"$\\frac{4}{9}$","correct":false,"explanation":"This uses the 4 blue counters as the numerator while retaining the simplified denominator incorrectly. There are 14 non-blue counters out of 18, giving $\\frac{7}{9}$."}]},{"id":"4c4b3528-c4ab-464e-bef2-a71cdc70395e","cardType":"basic","points":2,"front":"A probability table has red 0.25, blue 0.15 and yellow 0.40. What is the missing green probability?","back":"0.20","mcqOptions":[{"text":"0.20","correct":true,"explanation":"The probabilities must total 1. The known probabilities sum to $0.25 + 0.15 + 0.40 = 0.80$, leaving $1 - 0.80 = 0.20$."},{"text":"0.80","correct":false,"explanation":"This is the sum of the three known probabilities, not the missing probability. Subtracting that sum from 1 gives $1 - 0.80 = 0.20$."},{"text":"0.10","correct":false,"explanation":"This results from an arithmetic error when subtracting the known total from 1. Since the known probabilities total 0.80, the missing probability is 0.20."},{"text":"0.60","correct":false,"explanation":"This subtracts only the red and blue probabilities and overlooks yellow. All known probabilities must be included, so $1 - 0.25 - 0.15 - 0.40 = 0.20$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Probability","title":"Probability Flashcards","topbar":{"backHref":"/gcse/maths/edexcel/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"94c5a8ed-5472-4ded-a44d-f52afd4c0562","walled":"$L37"}],"$L38","$L39"]