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Subtracting the winning probability from 1 therefore gives $P(\\text{not win}) = 0.35$."},{"text":"$$1 - 0.65 = 0.45$","correct":false,"explanation":"This contains a subtraction error: $1.00 - 0.65$ is $0.35$, not $0.45$. Complementary branches must have probabilities that add exactly to 1."},{"text":"$$1 - 0.65 = 0.65$","correct":false,"explanation":"This incorrectly assumes that the two branches are equally likely. The complement is found by subtraction, so its probability is $1 - 0.65 = 0.35$."},{"text":"$$1 + 0.65 = 1.65$","correct":false,"explanation":"This adds the event probability instead of subtracting it from 1. A complement probability cannot exceed 1 and is $1 - 0.65 = 0.35$."}]},{"id":"f4570a8a-1d45-0daa-f597-4156ed4e9b80","cardType":"cloze","points":null,"front":"The sum of an event and its [ ] (not that event) always equals 1.","back":"The sum of an event and its complement (not that event) always equals 1.","mcqOptions":[{"text":"complement","correct":true,"explanation":"The complement of an event includes every outcome where that event does not happen. Together, an event and its complement cover all outcomes, so their probabilities sum to 1."},{"text":"intersection","correct":false,"explanation":"An intersection contains outcomes shared by two events; it is not the set of outcomes where an event does not occur. The required term is complement."},{"text":"union","correct":false,"explanation":"A union combines outcomes belonging to either of two events, rather than describing all outcomes outside one event. The term for “not that event” is complement."},{"text":"frequency","correct":false,"explanation":"Frequency records how often an outcome occurs and is not the opposite of an event. The event consisting of all outcomes where it does not happen is its complement."}]},{"id":"906a1c2a-55aa-fd84-4523-8b87184b7684","cardType":"cloze","points":null,"front":"The sum of an event and its complement (not that event) always equals [ ].","back":"The sum of an event and its complement (not that event) always equals 1.","mcqOptions":[{"text":"1","correct":true,"explanation":"An event and its complement cover every possible outcome without overlapping. Their probabilities therefore account for the whole sample space and must sum to 1."},{"text":"0","correct":false,"explanation":"A total of 0 would mean that neither the event nor its complement could occur, although together they cover every outcome. Their probabilities must sum to 1."},{"text":"0.5","correct":false,"explanation":"This assumes the event and its complement account for only half of all outcomes. Together they cover the complete sample space, so their probabilities always sum to 1."},{"text":"2","correct":false,"explanation":"This treats both probabilities as if each must equal 1. Although either probability can individually reach 1, an event and its complement must have a combined probability of 1."}]},{"id":"ff9db512-015e-3d41-2ca9-6e13951beef1","cardType":"basic","points":1,"front":"What is the rule for the sum of probabilities at any single split in a tree?","back":"They must add to 1.","mcqOptions":[{"text":"They must add to 1.","correct":true,"explanation":"The branches at one split represent all mutually exclusive next outcomes. Since one of those outcomes must occur, their probabilities must add to 1."},{"text":"They must add to 0.","correct":false,"explanation":"Probabilities at a split cannot total 0 because that would make every next outcome impossible. The branches cover all possible next outcomes, so they add to 1."},{"text":"They must multiply to 1.","correct":false,"explanation":"Multiplication is used along a complete route, not between alternative branches at one split. Probabilities on branches leaving the same point must add to 1."},{"text":"They must multiply to 0.","correct":false,"explanation":"This confuses combining outcomes at a split with calculating a route probability. Alternative branches are added and must total 1; multiplication is used along successive branches."}]},{"id":"03c1c15c-75a2-2ddd-aaa4-e2eef698644b","cardType":"basic","points":2,"front":"Define independent events in the context of probability trees.","back":"Events where the result of the first does not change the probabilities for the second.","mcqOptions":[{"text":"Events where the result of the first does not change the probabilities for the second.","correct":true,"explanation":"For independent events, knowing the first result gives no new information about the second. Consequently, the probabilities on the second-stage branches remain unchanged after every first result."},{"text":"Events where the result of the first always changes the probabilities for the second.","correct":false,"explanation":"This describes dependent events, where the first result affects what can happen next. For independent events, the probabilities for the second event do not change."},{"text":"Events where the result of the first determines exactly the result of the second.","correct":false,"explanation":"If the first result fixes the second result, the events are strongly dependent. Independent events require the second event’s probabilities to remain unaffected by the first result."},{"text":"Events where the result of the first equals the probability of the second.","correct":false,"explanation":"This incorrectly compares an outcome with a probability, which are different quantities. Independence means the first result does not alter the probabilities assigned to the second event."}]},{"id":"0a8a7ed4-f815-2dcd-bd69-5c06d04841f7","cardType":"basic","points":2,"front":"A bag contains 2 red and 6 blue counters. What labels go on the first set of branches?","back":"

$P(\\text{Red}) = \\frac{2}{8}$ (or $\\frac{1}{4}$)
$P(\\text{Blue}) = \\frac{6}{8}$ (or $\\frac{3}{4}$)

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

$P(\\text{Red}) = \\frac{2}{8}$ (or $\\frac{1}{4}$)
$P(\\text{Blue}) = \\frac{6}{8}$ (or $\\frac{3}{4}$)

","correct":true,"explanation":"There are 8 counters altogether. Probability is the number of counters of the required colour divided by 8, giving $2/8$ for red and $6/8$ for blue."},{"text":"

$P(\\text{Red}) = \\frac{6}{8}$ (or $\\frac{3}{4}$)
$P(\\text{Blue}) = \\frac{2}{8}$ (or $\\frac{1}{4}$)

","correct":false,"explanation":"This swaps the colour frequencies. Only 2 of the 8 counters are red and 6 are blue, so the correct probabilities are $2/8$ and $6/8$ respectively."},{"text":"

$P(\\text{Red}) = \\frac{2}{6}$ (or $\\frac{1}{3}$)
$P(\\text{Blue}) = \\frac{6}{6}$ (or $1$)

","correct":false,"explanation":"This uses the number of blue counters as the denominator instead of the total number of counters. Both probabilities need denominator 8, giving $2/8$ and $6/8$."},{"text":"

$P(\\text{Red}) = \\frac{2}{10}$ (or $\\frac{1}{5}$)
$P(\\text{Blue}) = \\frac{6}{10}$ (or $\\frac{3}{5}$)

","correct":false,"explanation":"This incorrectly adds the two colour counts to each numerator when forming the denominator. There are $2 + 6 = 8$ counters, so the denominator must be 8."}]},{"id":"a3315ef6-152e-70c8-24a1-be6f5b268cf4","cardType":"basic","points":1,"front":"How do you calculate the probability of one complete route (e.g. \"win and win\")?","back":"Multiply the probabilities along the branches of that route.","mcqOptions":[{"text":"Multiply the probabilities along the branches of that route.","correct":true,"explanation":"A complete route represents events happening together in sequence, such as win and win. The multiplication rule gives its probability by multiplying all branch probabilities along that route."},{"text":"Add the probabilities along the branches of that route.","correct":false,"explanation":"Adding is used to combine different alternative routes representing “or”. For events occurring along one route, representing “and”, multiply the branch probabilities instead."},{"text":"Subtract the probabilities along the branches of that route.","correct":false,"explanation":"Subtraction is used to find a complement from 1, not to combine successive events. A complete route uses the multiplication rule, so its branch probabilities are multiplied."},{"text":"Divide the probabilities along the branches of that route.","correct":false,"explanation":"Division is not the standard rule for finding the probability of successive events on one route. Since the route represents “and”, multiply its branch probabilities."}]},{"id":"2724d700-b35b-fc36-aefb-15bc585d379c","cardType":"cloze","points":null,"front":"In probability trees, the word \"and\" implies you should [ ] along branches, while \"or\" implies you should add different routes.","back":"In probability trees, the word \"and\" implies you should multiply along branches, while \"or\" implies you should add different routes.","mcqOptions":[{"text":"multiply","correct":true,"explanation":"Events joined by “and” occur along the same complete route, so their branch probabilities are multiplied. Probabilities of alternative routes joined by “or” are then added."},{"text":"add","correct":false,"explanation":"Adding along one route confuses the “and” and “or” rules. Probabilities are multiplied along branches for “and”, while separate alternative routes are added for “or”."},{"text":"divide","correct":false,"explanation":"Division is not used to combine branch probabilities for successive events. The word “and” identifies one complete route, whose probabilities must be multiplied together."},{"text":"subtract","correct":false,"explanation":"Subtraction is mainly used to calculate a complementary probability from 1. Events joined by “and” lie on one route, so their branch probabilities are multiplied."}]},{"id":"941f7a90-ed63-7795-50f0-d50aa16ab4ab","cardType":"cloze","points":null,"front":"In probability trees, the word \"and\" implies you should multiply along branches, while \"or\" implies you should [ ] different routes.","back":"In probability trees, the word \"and\" implies you should multiply along branches, while \"or\" implies you should add different routes.","mcqOptions":[{"text":"add","correct":true,"explanation":"Separate routes represent alternative ways an outcome can occur, so their probabilities are added after the probability of each complete route has been found by multiplication."},{"text":"multiply","correct":false,"explanation":"Multiplication is used for events joined by \"and\" along a single route. Alternative routes joined by \"or\" must instead be combined by adding their probabilities."},{"text":"subtract","correct":false,"explanation":"Subtraction is sometimes used for complements, such as finding an event from $1-P(\\text{not event})$. Different routes leading to the required outcome are added."},{"text":"divide","correct":false,"explanation":"Division is not the rule for combining alternative routes in a probability tree. Calculate each route by multiplication, then add the routes representing the required outcome."}]},{"id":"7905e236-1fea-7874-5a57-b053a1ddd705","cardType":"basic","points":2,"front":"Why should you NOT add probabilities when a question asks for \"both\" events to happen?","back":"Adding signifies \"one or the other\"; \"both\" requires following one full route via multiplication.","mcqOptions":[{"text":"Adding signifies \"one or the other\"; \"both\" requires following one full route via multiplication.","correct":true,"explanation":"Both events occur consecutively along one complete path through the tree. The branch probabilities on that path are therefore multiplied rather than added."},{"text":"Multiplying signifies \"one or the other\"; \"both\" requires combining separate routes via addition.","correct":false,"explanation":"This reverses the probability-tree rules. Multiplication follows consecutive events along one route, whereas addition combines separate routes that can each produce the required outcome."},{"text":"Adding is unsuitable only for dependent events; independent events can use direct addition for \"both\".","correct":false,"explanation":"Independence affects which probabilities appear on later branches, not the operation for both events. Their probabilities are still multiplied along the relevant complete route."},{"text":"Adding gives the complement of \"both\"; \"both\" requires subtracting route probabilities from 1.","correct":false,"explanation":"Complements are found by subtracting an opposite event from 1, but that is not the general meaning of both. Both events are found by multiplying along one route."}]},{"id":"db337236-fd2f-e6f4-4c5c-b3ec63600277","cardType":"cloze","points":null,"front":"A probability of [ ] represents an impossible event, while a probability of 1 represents a certain event.","back":"A probability of 0 represents an impossible event, while a probability of 1 represents a certain event.","mcqOptions":[{"text":"0","correct":true,"explanation":"Probability values range from 0 to 1 inclusive. A value of 0 means the event has no chance of occurring, so it is impossible."},{"text":"1","correct":false,"explanation":"A probability of 1 means the event is guaranteed, not impossible. An impossible event has probability 0 because it has no chance of occurring."},{"text":"0.5","correct":false,"explanation":"A probability of 0.5 represents an event with an equal chance of occurring or not occurring. An impossible event must have probability 0."},{"text":"-1","correct":false,"explanation":"Probabilities cannot be negative; they must lie between 0 and 1 inclusive. The valid probability representing an impossible event is 0."}]},{"id":"18cc0800-3d98-df54-2010-1cc925585aa5","cardType":"cloze","points":null,"front":"A probability of 0 represents an impossible event, while a probability of [ ] represents a certain event.","back":"A probability of 0 represents an impossible event, while a probability of 1 represents a certain event.","mcqOptions":[{"text":"1","correct":true,"explanation":"A probability of 1 means the event must occur, so it represents certainty. Probability values run from 0 for impossible to 1 for certain."},{"text":"0","correct":false,"explanation":"A probability of 0 represents an event that cannot occur, which is the opposite of certainty. A certain event has probability 1."},{"text":"0.5","correct":false,"explanation":"A probability of 0.5 indicates an even chance rather than certainty. An event that is guaranteed to occur has the maximum probability, 1."},{"text":"100","correct":false,"explanation":"This confuses a percentage with a probability written as a decimal. Certainty is 100%, but on the probability scale from 0 to 1 it is written as 1."}]},{"id":"191593fc-09a2-915b-d566-21512ade5dcb","cardType":"basic","points":2,"front":"A result can happen via two different routes. Outline the steps to find the total probability.","back":"

Multiply along each individual route
Add the resulting probabilities together

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

Multiply along each individual route
Add the resulting probabilities together

","correct":true,"explanation":"Multiplication finds the probability of all events along one complete route. Because either route produces the required result, the two route probabilities are then added."},{"text":"

Add along each individual route
Multiply the resulting probabilities together

","correct":false,"explanation":"This reverses the two probability-tree operations. Events along a route are joined by \"and\", so multiply them; alternative routes are joined by \"or\", so add them."},{"text":"

Multiply along each individual route
Subtract the resulting probabilities together

","correct":false,"explanation":"Multiplying along each route is correct, but subtracting would wrongly remove one valid route. Since either route gives the result, their probabilities must be added."},{"text":"

Add along each individual route
Add the resulting probabilities together

","correct":false,"explanation":"Adding at both stages treats consecutive events as alternatives. Multiply the branch probabilities along each route first, then add the probabilities of the two valid routes."}]},{"id":"aa5f4b35-02b2-8652-d7df-c072f3587c91","cardType":"basic","points":2,"front":"If $P(\\text{navy}) = \\frac{5}{9}$, what are the two specific routes for wearing different colours over two days (Navy or Grey)?","back":"

Navy then Grey
Grey then Navy

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

Navy then Grey
Grey then Navy

","correct":true,"explanation":"Different colours means one day must be Navy and the other Grey. Either order is possible, producing the two routes Navy-Grey and Grey-Navy."},{"text":"

Navy then Navy
Grey then Grey

","correct":false,"explanation":"These routes both produce matching colours on the two days. Different colours requires one Navy and one Grey, in either possible order."},{"text":"

Navy then Grey
Navy then Navy

","correct":false,"explanation":"The second route gives Navy on both days, so it does not satisfy different colours. The missing valid route is Grey then Navy."},{"text":"

Grey then Navy
Grey then Grey

","correct":false,"explanation":"The second route gives Grey on both days, so it does not satisfy different colours. The missing valid route is Navy then Grey."}]},{"id":"2ff2c460-73c7-177d-6b00-fd897ef819f0","cardType":"basic","points":2,"front":"What does the term at least one include when looking at two events?","back":"

First event only
Second event only
Both events

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

First event only
Second event only
Both events

","correct":true,"explanation":"At least one means one or more events occur. It therefore includes either event occurring alone as well as the case where both events occur."},{"text":"

First event only
Second event only
Neither event

","correct":false,"explanation":"Neither event means zero events occur, so it cannot be included in at least one. The third included possibility should be both events occurring."},{"text":"

First event only
Both events
Neither event

","correct":false,"explanation":"This wrongly includes zero events and omits the case where only the second event occurs. At least one includes first only, second only, and both."},{"text":"

Second event only
Both events
Neither event

","correct":false,"explanation":"This wrongly includes zero events and omits the case where only the first event occurs. At least one includes first only, second only, and both."}]},{"id":"9c189cf6-c6d7-e673-feb1-604ed0ee9c31","cardType":"cloze","points":null,"front":"The shortcut to calculate the probability of \"at least one\" success is $1 - P(\\text{\\text{[...]}})$.","back":"The shortcut to calculate the probability of \"at least one\" success is $1 - P(\\text{\\boxed{none}})$.","mcqOptions":[{"text":"none","correct":true,"explanation":"The complement of at least one success is no successes. Their probabilities total 1, so subtracting the probability of none gives the probability of at least one."},{"text":"one","correct":false,"explanation":"Subtracting the probability of one success does not generally leave the probability of at least one, because outcomes with multiple successes would be handled incorrectly. Subtract $P(\\text{none})$ instead."},{"text":"all","correct":false,"explanation":"All successes is not the complement of at least one success. The only outcomes excluded from at least one are those with no successes, so use $P(\\text{none})$."},{"text":"exactly one","correct":false,"explanation":"At least one includes exactly one and any larger number of successes. Its complement is therefore no successes, so the shortcut subtracts $P(\\text{none})$ from 1."}]},{"id":"5fa04937-68cb-c021-2a68-fa2e692437c2","cardType":"basic","points":3,"front":"The probability a train is late is 0.3. Use the shortcut to find the probability it is late on at least one of two days.","back":"

$P(\\text{not late}) = 0.7$
$P(\\text{none late}) = 0.7 \\times 0.7 = 0.49$
$1 - 0.49 = 0.51$

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

$P(\\text{not late}) = 0.7$
$P(\\text{none late}) = 0.7 \\times 0.7 = 0.49$
$1 - 0.49 = 0.51$

","correct":true,"explanation":"The shortcut finds the complement: calculate the probability that neither day is late, then subtract it from 1. This gives the probability of at least one late day."},{"text":"

$P(\\text{not late}) = 0.3$
$P(\\text{none late}) = 0.3 \\times 0.3 = 0.09$
$1 - 0.09 = 0.91$

","correct":false,"explanation":"This uses the late probability for the ‘not late’ branches, reversing the event. The correct complement is 0.7, giving $0.7 \\times 0.7 = 0.49$ and a final probability of 0.51."},{"text":"

$P(\\text{not late}) = 0.7$
$P(\\text{none late}) = 0.7 + 0.7 = 1.4$
$1 - 1.4 = -0.4$

","correct":false,"explanation":"Probabilities along a sequence of independent branches are multiplied, not added. Multiplying the two 0.7 branches gives 0.49, so the probability of at least one late day is 0.51."},{"text":"

$P(\\text{not late}) = 0.7$
$P(\\text{none late}) = 0.3 \\times 0.3 = 0.09$
$1 - 0.09 = 0.91$

","correct":false,"explanation":"The second line incorrectly uses 0.3, which is the probability of being late, for both ‘none late’ branches. Using 0.7 twice gives 0.49, so the required result is 0.51."}]},{"id":"c5643349-d0f6-e0fc-255d-9adf90fdcc31","cardType":"basic","points":1,"front":"In a three-outcome game, $P(\\text{win}) = 0.5$ and $P(\\text{draw}) = 0.2$. How do you find $P(\\text{lose})$?","back":"$$1 - 0.5 - 0.2 = 0.3$","mcqOptions":[{"text":"$$1 - 0.5 - 0.2 = 0.3$","correct":true,"explanation":"The three outcomes are exhaustive, so their probabilities add to 1. Subtracting the win and draw probabilities leaves $1 - 0.5 - 0.2 = 0.3$ for losing."},{"text":"$$1 - 0.5 + 0.2 = 0.7$","correct":false,"explanation":"The draw probability must also be subtracted because it is another non-losing outcome. Removing both known outcomes from 1 gives $1 - 0.5 - 0.2 = 0.3$."},{"text":"$$0.5 + 0.2 = 0.7$","correct":false,"explanation":"Adding the win and draw probabilities finds the chance of either winning or drawing, not losing. The remaining probability is found by subtracting that total from 1, giving 0.3."},{"text":"$$1 - 0.2 = 0.8$","correct":false,"explanation":"This only removes the draw outcome and forgets to remove wins as well. Since win and draw are both separate from losing, calculate $1 - 0.5 - 0.2 = 0.3$."}]},{"id":"ab2283a4-c47e-2ab3-292e-9a38ca9eceb4","cardType":"cloze","points":null,"front":"If an item is taken from a bag [ ], the probabilities on the second set of branches will change.","back":"If an item is taken from a bag without replacement, the probabilities on the second set of branches will change.","mcqOptions":[{"text":"without replacement","correct":true,"explanation":"Without replacement means the first item is not returned to the bag. The contents therefore change, so the probabilities on the second set of branches change too."},{"text":"with replacement","correct":false,"explanation":"With replacement means the first item is returned, leaving the bag’s contents unchanged. The probabilities on the second branches therefore stay the same, unlike the required phrase."},{"text":"at random","correct":false,"explanation":"Taking an item at random describes how it is selected, not whether the bag’s contents change afterwards. The relevant condition is that the item is taken without replacement."},{"text":"in pairs","correct":false,"explanation":"Taking items in pairs does not specifically state whether the first item is returned. The second branch probabilities change only when the item is taken without replacement."}]},{"id":"26f61c6e-4996-a03f-d517-270931450b06","cardType":"cloze","points":null,"front":"If an item is taken from a bag without replacement, the probabilities on the second set of branches will [ ].","back":"If an item is taken from a bag without replacement, the probabilities on the second set of branches will change.","mcqOptions":[{"text":"change","correct":true,"explanation":"Without replacement alters the contents of the bag after the first selection. Since the numbers of items change, the probabilities on the second set of branches change."},{"text":"stay constant","correct":false,"explanation":"The probabilities stay constant when an item is replaced after selection. Without replacement changes the bag’s composition, so the correct word is change."},{"text":"cancel","correct":false,"explanation":"Branch probabilities do not cancel when an item is removed. Removing an item changes the total and possibly the colour counts, so the probabilities change."},{"text":"double","correct":false,"explanation":"Removing one item does not generally double every second-branch probability. The probabilities may increase or decrease depending on the item removed, so they simply change."}]},{"id":"69142220-e60f-04c3-6933-d0e670654d8c","cardType":"basic","points":1,"front":"A player plays two rounds where $P(\\text{win}) = 0.5$. Calculate the probability they win both.","back":"$$0.5 \\times 0.5 = 0.25$","mcqOptions":[{"text":"$$0.5 \\times 0.5 = 0.25$","correct":true,"explanation":"The two rounds are treated as independent, so multiply the probability of winning each round. This gives $0.5 \\times 0.5 = 0.25$ for winning both."},{"text":"$$0.5 + 0.5 = 1$","correct":false,"explanation":"Adding probabilities finds neither the probability of both independent wins nor a valid result here. Probabilities along a route are multiplied, giving $0.5 \\times 0.5 = 0.25$."},{"text":"$$0.5 \\times 2 = 1$","correct":false,"explanation":"Multiplying by 2 counts the two rounds but does not calculate both events occurring. For consecutive independent wins, multiply the two branch probabilities to obtain 0.25."},{"text":"$$1 - (0.5 \\times 0.5) = 0.75$","correct":false,"explanation":"Subtracting the product from 1 finds the probability of not winning both, rather than winning both. The required probability is directly $0.5 \\times 0.5 = 0.25$."}]},{"id":"211297e4-0ec5-ca72-9025-6d47a3a5b284","cardType":"basic","points":1,"front":"What is the definition of a route in a probability tree?","back":"A complete path from the start of the tree to an end outcome.","mcqOptions":[{"text":"A complete path from the start of the tree to an end outcome.","correct":true,"explanation":"A route follows one branch at every stage, beginning at the start and finishing at one end outcome. It is therefore a complete path through the tree."},{"text":"A single branch from the start of the tree to an end outcome.","correct":false,"explanation":"A single branch represents one step between outcomes, not the whole sequence. A route includes every branch needed to form a complete path from start to an end outcome."},{"text":"A probability label beside each branch of the tree.","correct":false,"explanation":"Probability labels give the chance of individual branches, but they are not routes. A route is the complete sequence of branches from the tree’s start to an end outcome."},{"text":"A pair of outcomes at the same stage of the tree.","correct":false,"explanation":"Outcomes at one stage do not describe a full journey through the tree. A route must connect the start to an end outcome by following a complete path."}]},{"id":"d1d6b673-85f6-f1c7-7b1c-4d05189cdf48","cardType":"cloze","points":null,"front":"To find the probability of two independent events happening together, you [ ] their individual probabilities.","back":"To find the probability of two independent events happening together, you multiply their individual probabilities.","mcqOptions":[{"text":"multiply","correct":true,"explanation":"For independent events happening together, the probability of one event does not affect the other. Their individual probabilities are therefore multiplied to find the combined probability."},{"text":"add","correct":false,"explanation":"Adding probabilities is used for suitable ‘either-or’ outcomes, not independent events happening together. For two events both occurring, multiply their individual probabilities."},{"text":"subtract","correct":false,"explanation":"Subtraction is commonly used with complements, such as finding an event from ‘not the event’. Two independent events occurring together require multiplication instead."},{"text":"divide","correct":false,"explanation":"Dividing probabilities does not combine independent events occurring together. The probability along a route is found by multiplying the probabilities of its branches."}]},{"id":"29c88c5c-4cf6-2cbd-7848-e9355f08fb58","cardType":"basic","points":1,"front":"If $P(\\text{win badminton}) = \\frac{3}{5}$ and $P(\\text{win tennis}) = \\frac{7}{10}$, calculate $P(\\text{win both})$.","back":"$$\\frac{3}{5} \\times \\frac{7}{10} = \\frac{21}{50}$","mcqOptions":[{"text":"$$\\frac{3}{5} \\times \\frac{7}{10} = \\frac{21}{50}$","correct":true,"explanation":"Winning both independent sports requires both events to occur, so multiply their probabilities. The product is $\\frac{21}{50}$ after multiplying the numerators and denominators."},{"text":"$$\\frac{3}{5} + \\frac{7}{10} = \\frac{13}{10}$","correct":false,"explanation":"Adding finds a combined ‘either’ quantity and produces a value above 1, which cannot be a probability. For winning both, multiply to get $\\frac{21}{50}$."},{"text":"$$\\frac{3}{5} \\times \\frac{7}{10} = \\frac{21}{100}$","correct":false,"explanation":"The numerators are multiplied correctly, but the denominator calculation is wrong: $5 \\times 10 = 50$, not 100. Therefore the product is $\\frac{21}{50}$."},{"text":"$$\\frac{3}{5} \\times \\frac{7}{10} = \\frac{10}{15}$","correct":false,"explanation":"This adds the numerators and denominators rather than multiplying the fractions. Multiplying corresponding parts gives $\\frac{3 \\times 7}{5 \\times 10} = \\frac{21}{50}$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Probability Trees","title":"Probability Trees Flashcards","topbar":{"backHref":"/gcse/maths/ocr/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"470732b0-5bcc-4cf8-9edf-17310fe4be59","walled":"$L38"}],"$L39","$L3a"]