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occurs."},{"text":"0","correct":false,"explanation":"A total of 0 would mean that none of the possible outcomes can occur. Since the branches cover all outcomes at the split, their probabilities must add to 1."},{"text":"0.5","correct":false,"explanation":"The total is not automatically 0.5 when there are two branches; individual probabilities may vary. All branches from a single split must add to 1."},{"text":"100","correct":false,"explanation":"This confuses a probability with a percentage value. Probabilities written as decimals or fractions add to 1, while probabilities written as percentages add to 100%."}]},{"id":"38829a75-9080-4ba2-a683-eb8d1823a90b","cardType":"basic","points":1,"front":"If the probability of an event happening is $p$, what is the probability of the event not happening?","back":"$$1 - p$","mcqOptions":[{"text":"$$1 - p$","correct":true,"explanation":"An event and its complement cover all possible outcomes, so their probabilities add to 1. Therefore, the probability that the event does not happen is $1-p$."},{"text":"$$1 + p$","correct":false,"explanation":"Adding $p$ can produce a value greater than 1, which is impossible for a probability. The complementary probability is found by subtracting $p$ from 1."},{"text":"$$p - 1$","correct":false,"explanation":"Reversing the subtraction usually produces a negative value, which cannot be a probability. The probability of the event not happening is $1-p$."},{"text":"$$rac{1}{p}$","correct":false,"explanation":"Taking the reciprocal does not find a complementary probability and may give a result greater than 1. The event and its complement must sum to 1, giving $1-p$."}]},{"id":"98f4958d-ea8c-4fbb-9e4f-81bea2a1d912","cardType":"cloze","points":null,"front":"The general formula for the probability of an event is: $\\text{[...]}$","back":"The general formula for the probability of an event is: $\\boxed{\\frac{\\text{number of successful outcomes}}{\\text{total number of outcomes}}}$","mcqOptions":[{"text":"$$\\frac{\\text{number of successful outcomes}}{\\text{total number of outcomes}}$","correct":true,"explanation":"For equally likely outcomes, probability is calculated by dividing the number of outcomes that satisfy the event by the total number of possible outcomes."},{"text":"$$\\frac{\\text{total number of outcomes}}{\\text{number of successful outcomes}}$","correct":false,"explanation":"This reverses the fraction and can produce a value greater than 1. Probability is successful outcomes divided by the total number of outcomes."},{"text":"$$\\frac{\\text{number of unsuccessful outcomes}}{\\text{total number of outcomes}}$","correct":false,"explanation":"This calculates the probability of the event not occurring by counting unsuccessful outcomes. The required probability uses the number of successful outcomes as the numerator."},{"text":"$$\\frac{\\text{number of successful outcomes}}{\\text{number of unsuccessful outcomes}}$","correct":false,"explanation":"This compares successful outcomes only with unsuccessful ones, rather than with every possible outcome. The denominator must be the total number of outcomes."}]},{"id":"faf5c64e-1c77-43d5-b97e-6472b5e0b08e","cardType":"basic","points":1,"front":"When is a probability tree diagram preferred over a simple probability calculation?","back":"When there is a sequence of events or more than one stage.","mcqOptions":[{"text":"When there is a sequence of events or more than one stage.","correct":true,"explanation":"A probability tree clearly represents successive stages and shows how each event branches into later possibilities, making multi-stage probability calculations easier to organise."},{"text":"When there is a single event or only one possible outcome.","correct":false,"explanation":"A single event or one possible outcome does not usually need a branching diagram. Probability trees are preferred for a sequence of events or more than one stage."},{"text":"When there are equally likely outcomes or no later stages.","correct":false,"explanation":"Equally likely outcomes alone do not require a tree, especially when there are no subsequent stages. Trees are useful when events occur in sequence across multiple stages."},{"text":"When there is a single calculation or only one stage.","correct":false,"explanation":"A one-stage calculation can normally be completed directly without a tree. A probability tree is preferred when there is a sequence of events or more than one stage."}]},{"id":"ddf0cc65-badf-419f-b002-fb0f8865423b","cardType":"basic","points":1,"front":"How do you calculate the probability of a single complete path through a probability tree?","back":"Multiply the probabilities along the branches of that path.","mcqOptions":[{"text":"Multiply the probabilities along the branches of that path.","correct":true,"explanation":"A complete path represents events occurring together in sequence, so the multiplication rule is used to find the probability of following every branch on that path."},{"text":"Add the probabilities along the branches of that path.","correct":false,"explanation":"Adding is used to combine probabilities from separate alternative paths, not branches within one path. For a single complete path, multiply the branch probabilities."},{"text":"Subtract the probabilities along the branches of that path.","correct":false,"explanation":"Subtraction is associated with complements, such as finding a probability from 1. A complete path represents successive events, so its branch probabilities are multiplied."},{"text":"Divide the probabilities along the branches of that path.","correct":false,"explanation":"Division does not combine the successive events represented by a path. To find the probability that all stages on that path occur, multiply their branch probabilities."}]},{"id":"60aea8a3-6de8-4308-8123-04a61a4067a1","cardType":"basic","points":2,"front":"What defines two events as being independent?","back":"The outcome of the first event does not change the probabilities of the second event.","mcqOptions":[{"text":"The outcome of the first event does not change the probabilities of the second event.","correct":true,"explanation":"Independent events do not influence one another, so knowing the outcome of the first event leaves the probabilities at the second stage unchanged."},{"text":"The outcome of the first event does change the probabilities of the second event.","correct":false,"explanation":"This describes dependent events, where the first result affects later probabilities. For independent events, the first outcome does not change the probabilities of the second event."},{"text":"The outcome of the first event determines exactly the outcome of the second event.","correct":false,"explanation":"If the first outcome fixes the second, the events are strongly dependent rather than independent. Independence means the first event does not change second-event probabilities."},{"text":"The outcome of the first event must match the outcome of the second event.","correct":false,"explanation":"Matching outcomes are not a requirement for independence and may indicate a relationship between events. Independent events are defined by one outcome not changing the other’s probabilities."}]},{"id":"92a16435-22df-462d-802d-acd157de65aa","cardType":"cloze","points":null,"front":"If an item is taken from a bag and not replaced, the probabilities for the next stage of the tree will [ ].","back":"If an item is taken from a bag and not replaced, the probabilities for the next stage of the tree will change.","mcqOptions":[{"text":"change","correct":true,"explanation":"Without replacement, the bag’s contents and total number of items are altered. This changes the fractions used for the probabilities at the next stage."},{"text":"stay constant","correct":false,"explanation":"Probabilities stay constant when replacement restores the original contents. Without replacement, both the available outcomes and total number of items change, so the next probabilities change."},{"text":"remain equal","correct":false,"explanation":"The next-stage probabilities are not necessarily equal because one item has been removed. Without replacement, the composition of the bag changes, so the probabilities change."},{"text":"return to zero","correct":false,"explanation":"Removing an item does not make all later outcomes impossible. The remaining items still have positive probabilities, but those probabilities change because the bag’s contents have changed."}]},{"id":"d3546ece-a9d1-4211-9020-8a90fb99b3e7","cardType":"basic","points":2,"front":"If a question asks for the probability of \"different colours\" over two stages, how do you combine the relevant paths?","back":"Add the individual path probabilities together.","mcqOptions":[{"text":"Add the individual path probabilities together.","correct":true,"explanation":"Different colours can occur through alternative complete paths, such as red then blue or blue then red. Add these mutually exclusive path probabilities to obtain the overall probability."},{"text":"Multiply the individual path probabilities together.","correct":false,"explanation":"Multiplication finds the probability along each individual path, not the total across alternative paths. After calculating each different-colour path, add their probabilities together."},{"text":"Subtract the individual path probabilities from one another.","correct":false,"explanation":"The relevant paths are alternative ways to obtain different colours, so neither should be removed from the other. Calculate each path probability and add them together."},{"text":"Divide the individual path probabilities by one another.","correct":false,"explanation":"Division does not combine alternative outcomes on a probability tree. Different-colour paths are separate mutually exclusive possibilities, so their individual probabilities must be added together."}]},{"id":"bec9274c-f1d6-4347-9a28-24496a3799f0","cardType":"basic","points":2,"front":"What is the most efficient way to calculate the probability of at least one success over multiple trials?","back":"Calculate $1 - P(\\text{none})$.","mcqOptions":[{"text":"Calculate $1 - P(\\text{none})$.","correct":true,"explanation":"The event “at least one success” is the complement of no successes. Therefore, subtracting the probability of no successes from 1 gives the required probability."},{"text":"Calculate $P(\\text{none})$.","correct":false,"explanation":"This finds the probability of no successes, not at least one success. The required complementary probability is calculated using $1 - P(\\text{none})$."},{"text":"Calculate $1 - P(\\text{exactly one})$.","correct":false,"explanation":"Exactly one success is not the complement of at least one success, because two or more successes are also included. The complement is no successes, so use $1 - P(\\text{none})$."},{"text":"Add the probability of success on every trial.","correct":false,"explanation":"Simply adding individual success probabilities can double-count outcomes and may produce a value above 1. The efficient method is to calculate $1 - P(\\text{none})$."}]},{"id":"7b658fbc-cf6e-4dad-91ac-2c123bac881f","cardType":"cloze","points":null,"front":"The [ ] of an event refers to the probability that the event does not happen.","back":"The complement of an event refers to the probability that the event does not happen.","mcqOptions":[{"text":"complement","correct":true,"explanation":"An event and its complement cover all possible outcomes without overlapping. Their probabilities therefore add to 1, so the complement represents the event not happening."},{"text":"intersection","correct":false,"explanation":"An intersection describes two or more events occurring together, usually represented by AND. The probability that an event does not happen is called its complement."},{"text":"union","correct":false,"explanation":"A union describes outcomes belonging to either of two events and is associated with OR. The probability that a particular event does not happen is its complement."},{"text":"sample space","correct":false,"explanation":"The sample space is the set of all possible outcomes, rather than the probability of an event not occurring. That non-occurrence is represented by the complement."}]},{"id":"c8414048-5ef9-4c39-a295-73ba13c3d3b2","cardType":"cloze","points":null,"front":"The complement of an event refers to the probability that the event [ ].","back":"The complement of an event refers to the probability that the event does not happen.","mcqOptions":[{"text":"does not happen","correct":true,"explanation":"The complement contains every outcome outside the original event. It therefore represents the probability that the specified event does not occur."},{"text":"does happen","correct":false,"explanation":"This describes the original event rather than its complement. The complement refers to all outcomes for which the event does not happen."},{"text":"must happen","correct":false,"explanation":"An event that must happen is a certain event with probability 1, not a general complement. A complement describes the event not happening."},{"text":"cannot happen","correct":false,"explanation":"An event that cannot happen is an impossible event with probability 0. A complement instead gives the probability that the original event does not happen."}]},{"id":"5721cc74-7aae-41ae-94b1-980290e7882b","cardType":"basic","points":1,"front":"A bag contains 5 red and 3 blue counters. Why is the probability of picking red $\\frac{5}{8}$ and not $\\frac{5}{3}$?","back":"Probabilities must be calculated using the total number of outcomes ($5 + 3 = 8$).","mcqOptions":[{"text":"Probabilities must be calculated using the total number of outcomes ($5 + 3 = 8$).","correct":true,"explanation":"There are eight counters altogether, and five of these are red. Probability is favourable outcomes divided by total outcomes, giving $P(\\text{red})=\\frac{5}{8}$."},{"text":"Probabilities must be calculated using the number of blue outcomes ($3$).","correct":false,"explanation":"Using only the three blue counters as the denominator confuses unfavourable outcomes with total outcomes. The denominator must be all eight counters, giving $\\frac{5}{8}$."},{"text":"Probabilities must be calculated using the number of red outcomes ($5$).","correct":false,"explanation":"Using the five red counters as both numerator and denominator would incorrectly give probability 1. The denominator is the total of eight counters, so the probability is $\\frac{5}{8}$."},{"text":"Probabilities must be calculated using the product of the outcomes ($5 \\times 3 = 15$).","correct":false,"explanation":"Multiplying the red and blue counts does not give the number of possible single-counter outcomes. Add the counts to get eight, making the probability $\\frac{5}{8}$."}]},{"id":"de278e02-2095-4e50-afe6-f5d0c99df2fe","cardType":"basic","points":2,"front":"Calculate the probability of picking two blue counters if Bag A has $P(\\text{blue}) = \\frac{2}{5}$ and Bag B has $P(\\text{blue}) = \\frac{3}{5}$.","back":"$$\\frac{2}{5} \\times \\frac{3}{5} = \\frac{6}{25}$","mcqOptions":[{"text":"$$\\frac{2}{5} \\times \\frac{3}{5} = \\frac{6}{25}$","correct":true,"explanation":"Both blue selections must occur, so the probabilities along the relevant tree path are multiplied. Thus $\\frac{2}{5}\\times\\frac{3}{5}=\\frac{6}{25}$."},{"text":"$$\\frac{2}{5} + \\frac{3}{5} = 1$","correct":false,"explanation":"Adding is used to combine alternative paths, not two events that must both occur. For blue from A and blue from B, multiply to obtain $\\frac{6}{25}$."},{"text":"$$\\frac{2}{5} \\times \\frac{3}{5} = \\frac{6}{10}$","correct":false,"explanation":"The numerators have been multiplied but the denominators have been added. Multiplying fractions requires multiplying both parts, so the denominator is $5\\times5=25$ and the answer is $\\frac{6}{25}$."},{"text":"$$\\frac{2}{5} \\times \\frac{2}{5} = \\frac{4}{25}$","correct":false,"explanation":"This incorrectly uses Bag A’s blue probability for both selections. Bag B has blue probability $\\frac{3}{5}$, so the correct product is $\\frac{2}{5}\\times\\frac{3}{5}=\\frac{6}{25}$."}]},{"id":"b3b6eb1c-cb08-4fdc-ba70-371f957869a4","cardType":"cloze","points":null,"front":"In probability tree problems, AND situations usually require [ ] while OR situations require addition.","back":"In probability tree problems, AND situations usually require multiplication while OR situations require addition.","mcqOptions":[{"text":"multiplication","correct":true,"explanation":"An AND situation follows a complete path through the probability tree. Multiplying the branch probabilities gives the joint probability that every event on that path occurs."},{"text":"addition","correct":false,"explanation":"Addition usually combines alternative, mutually exclusive paths associated with OR. When events on one path must both happen, their branch probabilities are combined by multiplication."},{"text":"subtraction","correct":false,"explanation":"Subtracting probabilities is generally used for complements or differences, not for events that must occur together. AND probabilities along a tree path are found by multiplication."},{"text":"division","correct":false,"explanation":"Division can be relevant when finding conditional probabilities from known values, but it does not combine branches along a path. AND situations normally require multiplication."}]},{"id":"b4670be7-a2f2-4258-a64f-abef1133f084","cardType":"cloze","points":null,"front":"In probability tree problems, AND situations usually require multiplication while OR situations require [ ].","back":"In probability tree problems, AND situations usually require multiplication while OR situations require addition.","mcqOptions":[{"text":"addition","correct":true,"explanation":"An OR situation combines alternative outcomes or complete paths. When those alternatives are mutually exclusive, their probabilities are added to find the overall probability."},{"text":"multiplication","correct":false,"explanation":"Multiplication combines successive events along one path and therefore represents AND. Alternative mutually exclusive paths representing OR are combined using addition."},{"text":"subtraction","correct":false,"explanation":"Subtraction is useful for complements, such as finding $1-P(A)$, but it does not combine alternative successful paths. OR situations normally require addition."},{"text":"division","correct":false,"explanation":"Division is not the standard operation for combining alternative outcomes in a probability tree. The probabilities of mutually exclusive OR paths are added together."}]},{"id":"a8c70a90-97c4-4a6a-8d38-c6b4fe59a47e","cardType":"basic","points":1,"front":"If $P(\\text{Win}) = 0.5$ and $P(\\text{Draw}) = 0.2$ for a game, what is the probability of losing?","back":"$$1 - 0.5 - 0.2 = 0.3$","mcqOptions":[{"text":"$$1 - 0.5 - 0.2 = 0.3$","correct":true,"explanation":"Winning, drawing and losing are mutually exclusive and exhaustive, so their probabilities total 1. Subtracting $0.5$ and $0.2$ leaves a losing probability of $0.3$."},{"text":"$$0.5 + 0.2 = 0.7$","correct":false,"explanation":"Adding $0.5$ and $0.2$ gives the probability of winning or drawing, not losing. Losing is the complement, so its probability is $1-0.7=0.3$."},{"text":"$$1 - 0.5 = 0.5$","correct":false,"explanation":"This subtracts only the winning probability and incorrectly includes draws with losses. Both winning and drawing must be removed from 1, leaving $0.3$."},{"text":"$$1 - 0.2 = 0.8$","correct":false,"explanation":"This subtracts only the drawing probability and incorrectly includes wins with losses. Subtract both $0.5$ and $0.2$ from 1 to obtain $0.3$."}]},{"id":"93e3c581-1f37-4e3f-bc0d-aa4398d12b3a","cardType":"basic","points":3,"front":"The probability of a train being late is 0.3. Calculate the probability it is late on at least one of two days.","back":"

$P(\\text{not late}) = 0.7$
$P(\\text{never late}) = 0.7 \\times 0.7 = 0.49$
$1 - 0.49 = 0.51$

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

$P(\\text{not late}) = 0.7$
$P(\\text{never late}) = 0.7 \\times 0.7 = 0.49$
$1 - 0.49 = 0.51$

","correct":true,"explanation":"The complement of being late at least once is never being late. Assuming the days are independent, this has probability $0.7^2=0.49$, so the required probability is $1-0.49=0.51$."},{"text":"

$P(\\text{late}) = 0.3$
$P(\\text{late twice}) = 0.3 \\times 0.3 = 0.09$
$P(\\text{at least once}) = 0.09$

","correct":false,"explanation":"Multiplying $0.3$ by $0.3$ finds the probability that the train is late on both days, not at least one day. Using the complement gives $1-0.7^2=0.51$."},{"text":"

$P(\\text{late}) = 0.3$
$P(\\text{late on either day}) = 0.3 + 0.3 = 0.6$
$P(\\text{at least once}) = 0.6$

","correct":false,"explanation":"Adding the two daily probabilities double-counts the outcome in which the train is late on both days. The complement method avoids this overlap and gives $1-0.7^2=0.51$."},{"text":"

$P(\\text{late}) = 0.3$
$P(\\text{late twice}) = 0.3 \\times 0.3 = 0.09$
$1 - 0.09 = 0.91$

","correct":false,"explanation":"Subtracting the probability of two late days finds the probability that the train is not late on both days. The required complement is never late, so the answer is $1-0.7^2=0.51$."}]},{"id":"82603cb9-ce79-4cde-a0de-3b4328e4f6a6","cardType":"basic","points":1,"front":"What should you check at every \"split\" in a probability tree before starting calculations?","back":"Check that the probabilities of the branches coming out add to 1.","mcqOptions":[{"text":"Check that the probabilities of the branches coming out add to 1.","correct":true,"explanation":"Branches leaving the same split represent all mutually exclusive possible outcomes. Their probabilities must therefore total 1, allowing a missing branch probability to be found by subtraction."},{"text":"Check that the probabilities of the branches coming out multiply to 1.","correct":false,"explanation":"Multiplication is used along a path through successive events, not between branches at one split. Probabilities leaving any single split must add to 1 because they cover all outcomes."},{"text":"Check that the probabilities of the branches coming out add to 0.","correct":false,"explanation":"A total of 0 would mean that none of the outcomes can occur. Since the branches represent every possible outcome at that split, their probabilities must add to 1."},{"text":"Check that the probabilities of the branches coming out are all equal.","correct":false,"explanation":"Branches are equal only when the outcomes are equally likely, which is not generally true. The universal check is that all probabilities leaving the same split add to 1."}]},{"id":"ed4b8921-774d-4235-ae37-98df78d18d50","cardType":"cloze","points":null,"front":"For two trials, the opposite (complement) of \"at least one success\" is [ ].","back":"For two trials, the opposite (complement) of \"at least one success\" is no successes.","mcqOptions":[{"text":"no successes","correct":true,"explanation":"The event “at least one success” includes exactly one success and two successes. Its complement contains the only remaining possibility, which is obtaining no successes in the two trials."},{"text":"two successes","correct":false,"explanation":"Two successes is included within “at least one success”, so it cannot be its complement. The opposite event must exclude every success outcome and is therefore no successes."},{"text":"one success","correct":false,"explanation":"Exactly one success is one of the outcomes covered by “at least one success”, rather than its opposite. The complement is no successes across the two trials."},{"text":"at least one failure","correct":false,"explanation":"At least one failure can occur alongside one success, so it overlaps with “at least one success”. The true complement contains no overlap and is no successes."}]},{"id":"052ddf15-9253-4d24-b5b3-051ea99aa12a","cardType":"basic","points":3,"front":"Sofia wears a blue scarf with $P = \\frac{3}{7}$ or a white scarf. Calculate the probability she wears different colours on Monday and Tuesday.","back":"

$P(\\text{Blue, White}) = \\frac{3}{7} \\times \\frac{4}{7} = \\frac{12}{49}$
$P(\\text{White, Blue}) = \\frac{4}{7} \\times \\frac{3}{7} = \\frac{12}{49}$
$\\frac{12}{49} + \\frac{12}{49} = \\frac{24}{49}$

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

$P(\\text{Blue, White}) = \\frac{3}{7} \\times \\frac{4}{7} = \\frac{12}{49}$
$P(\\text{White, Blue}) = \\frac{4}{7} \\times \\frac{3}{7} = \\frac{12}{49}$
$\\frac{12}{49} + \\frac{12}{49} = \\frac{24}{49}$

","correct":true,"explanation":"Different colours can occur in two mutually exclusive orders: blue then white, or white then blue. Each has probability $\\frac{12}{49}$, so adding them gives $\\frac{24}{49}$."},{"text":"

$P(\\text{Blue, Blue}) = \\frac{3}{7} \\times \\frac{3}{7} = \\frac{9}{49}$
$P(\\text{White, White}) = \\frac{4}{7} \\times \\frac{4}{7} = \\frac{16}{49}$
$\\frac{9}{49} + \\frac{16}{49} = \\frac{25}{49}$

","correct":false,"explanation":"These paths describe wearing the same colour on both days, the complement of the required event. The two different-colour paths total $\\frac{12}{49}+\\frac{12}{49}=\\frac{24}{49}$."},{"text":"

$P(\\text{Blue, White}) = \\frac{3}{7} \\times \\frac{4}{7} = \\frac{12}{49}$
$P(\\text{White, Blue}) = \\frac{4}{7} \\times \\frac{3}{7} = \\frac{12}{49}$
$\\frac{12}{49} \\times \\frac{12}{49} = \\frac{144}{2401}$

","correct":false,"explanation":"Multiplication is used along each path, but alternative mutually exclusive paths must be added. Adding blue-white and white-blue gives $\\frac{12}{49}+\\frac{12}{49}=\\frac{24}{49}$."},{"text":"

$P(\\text{Blue, White}) = \\frac{3}{7} \\times \\frac{4}{6} = \\frac{2}{7}$
$P(\\text{White, Blue}) = \\frac{4}{7} \\times \\frac{3}{6} = \\frac{2}{7}$
$\\frac{2}{7} + \\frac{2}{7} = \\frac{4}{7}$

","correct":false,"explanation":"Changing the denominator to 6 incorrectly treats the scarf choice as sampling without replacement. The daily probabilities remain $\\frac{3}{7}$ and $\\frac{4}{7}$, giving $\\frac{24}{49}$ overall."}]},{"id":"9f2b683e-f327-476c-8f5c-0d7a4cb0e39d","cardType":"basic","points":2,"front":"What do probabilities of 0 and 1 represent respectively?","back":"

0 = impossible
1 = certain

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

0 = impossible
1 = certain

","correct":true,"explanation":"Probability 0 means an event cannot occur, while probability 1 means it must occur. All probabilities lie between these endpoints, inclusive."},{"text":"

0 = certain
1 = impossible

","correct":false,"explanation":"This reverses the probability scale. A probability of 0 represents an impossible event, whereas a probability of 1 represents an event that is certain to occur."},{"text":"

0 = unlikely
1 = likely

","correct":false,"explanation":"“Unlikely” and “likely” describe probabilities near the endpoints, not the endpoints themselves. Exactly 0 means impossible, and exactly 1 means certain."},{"text":"

0 = even chance
1 = certain

","correct":false,"explanation":"An even chance has probability $0.5$, not 0. Probability 0 means impossible, while probability 1 correctly represents certainty."}]},{"id":"06996e2b-e02a-4b60-9c7a-818f6554a6b8","cardType":"basic","points":2,"front":"Amir wins table tennis with $P = 0.6$ and darts with $P = 0.75$. Calculate the probability he wins both games.","back":"$$0.6 \\times 0.75 = 0.45$","mcqOptions":[{"text":"$$0.6 \\times 0.75 = 0.45$","correct":true,"explanation":"Winning both games means following the win branch for table tennis and the win branch for darts. Multiplying the independent probabilities gives $0.6\\times0.75=0.45$."},{"text":"$$0.6 + 0.75 = 1.35$","correct":false,"explanation":"Adding the probabilities does not find the chance of both events and produces a value above 1, which is impossible for a probability. Multiply instead to obtain $0.45$."},{"text":"$$0.6 \\times 0.75 = 0.045$","correct":false,"explanation":"The multiplication has a decimal-place error: $6\\times75=450$, with three decimal places in total, giving $0.450$. Therefore, the probability is $0.45$."},{"text":"$$1 - (0.4 \\times 0.25) = 0.9$","correct":false,"explanation":"This subtracts the probability of losing both games, so it finds winning at least one game. Winning both requires multiplying the two win probabilities, giving $0.6\\times0.75=0.45$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Probability Trees","title":"Probability Trees Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"bd51f121-c6e4-4963-ad53-0c4aefb074ba","walled":"$L39"}],"$L3a","$L3b"]