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1.","mcqOptions":[{"text":"1","correct":true,"explanation":"Probabilities for all possible outcomes from one point must total 1 because those branches collectively represent every outcome that can occur next."},{"text":"0","correct":false,"explanation":"A total of 0 would mean that none of the possible outcomes can occur. The branches must instead sum to 1, representing certainty."},{"text":"2","correct":false,"explanation":"This confuses the number of branches with their total probability. Regardless of how many branches there are, their probabilities must sum to 1."},{"text":"0.5","correct":false,"explanation":"Probabilities may be written as percentages, but 50% is not the required total. All branches from the same point must sum to 1, or 100%."}]},{"id":"ff56142c-4b65-8d3f-5eaa-cf631519bb8d","cardType":"basic","points":1,"front":"What does the notation $P(B \\mid A)$ represent?","back":"The conditional probability of B occurring given that A has already occurred.","mcqOptions":[{"text":"The conditional probability of B occurring given that A has already occurred.","correct":true,"explanation":"The vertical bar means “given that”, so this notation describes the probability that B occurs under the condition that A has already occurred."},{"text":"The conditional probability of A occurring given that B has already occurred.","correct":false,"explanation":"This reverses the condition and describes $P(A \\mid B)$. In $P(B \\mid A)$, B is the event being considered after A has already occurred."},{"text":"The combined probability of A and B both occurring.","correct":false,"explanation":"The notation for both events occurring is $P(A \\cap B)$ or $P(A \\text{ and } B)$. The vertical bar instead indicates B occurring given A."},{"text":"The total probability of A or B occurring independently.","correct":false,"explanation":"The probability of either event is represented using a union, such as $P(A \\cup B)$. Here the vertical bar specifies the condition that A has occurred."}]},{"id":"9ecd8596-a9a8-4964-afde-3deaa596c208","cardType":"basic","points":1,"front":"How do you calculate the probability of a specific sequence of events (a single route) in a tree diagram?","back":"Multiply the probabilities along the branches.","mcqOptions":[{"text":"Multiply the probabilities along the branches.","correct":true,"explanation":"A route represents successive events that must all occur, so the multiplication rule is used to combine the probabilities on its branches."},{"text":"Add the probabilities along the branches.","correct":false,"explanation":"Adding is used to combine separate mutually exclusive routes, not successive branches on one route. For a specific sequence, multiply the branch probabilities."},{"text":"Subtract the probabilities along the branches.","correct":false,"explanation":"Subtracting branch probabilities does not represent successive events. A specific sequence requires every event on its route to occur, so multiply the probabilities."},{"text":"Divide the probabilities along the branches.","correct":false,"explanation":"Dividing the branch probabilities is not the tree-diagram rule for a sequence. Since all events along the route must occur, their probabilities are multiplied."}]},{"id":"ce9de868-4965-28fc-9c9c-02f8e00d5502","cardType":"cloze","points":null,"front":"The rule for finding the probability of a sequence of events is $$P(A \\text{ and } B) = \\text{[...]}$$.","back":"The rule for finding the probability of a sequence of events is $$P(A \\text{ and } B) = \\boxed{P(A) \\times P(B \\mid A)}$$.","mcqOptions":[{"text":"$$P(A) \\times P(B \\mid A)$","correct":true,"explanation":"For A and B to occur in sequence, multiply the probability of A by the conditional probability that B occurs after A has occurred."},{"text":"$$P(A) + P(B \\mid A)$","correct":false,"explanation":"Adding these probabilities combines alternatives rather than successive events. The probability of the sequence is $P(A) \\times P(B \\mid A)$."},{"text":"$$P(A) \\times P(B)$","correct":false,"explanation":"This incorrectly treats B as unaffected by A. The general multiplication rule uses B’s conditional probability, giving $P(A) \\times P(B \\mid A)$."},{"text":"$$P(A) \\times P(A \\mid B)$","correct":false,"explanation":"This reverses which event is conditioned on while retaining $P(A)$. The matching rule is $P(A) \\times P(B \\mid A)$."}]},{"id":"a6ccb719-9ed4-854f-75e7-3fb2d581e41a","cardType":"basic","points":2,"front":"In a \"without replacement\" scenario, what two things change for the second choice?","back":"

The total number of items (denominator) decreases by 1.
The number of items of the chosen category (numerator) decreases by 1.

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

The total number of items (denominator) decreases by 1.
The number of items of the chosen category (numerator) decreases by 1.

","correct":true,"explanation":"Because the selected item is not returned, one fewer item remains overall and one fewer remains in the category that was selected."},{"text":"

The total number of items (denominator) stays the same.
The number of items of the chosen category (numerator) stays the same.

","correct":false,"explanation":"This describes replacement rather than selection without replacement. The denominator must decrease by 1, and the chosen category’s numerator must also decrease by 1."},{"text":"

The total number of items (denominator) increases by 1.
The number of items of the chosen category (numerator) increases by 1.

","correct":false,"explanation":"Removing an item reduces rather than increases the available quantities. Both the total and the count for the selected category decrease by 1."},{"text":"

The total number of items (denominator) stays the same.
The number of items of the chosen category (numerator) decreases by 1.

","correct":false,"explanation":"The denominator cannot stay unchanged because one item has been removed. It decreases by 1, as does the numerator for the category selected first."}]},{"id":"735d69e7-ba6e-0bb6-5585-b9eb0d929e78","cardType":"basic","points":1,"front":"When should you add separate routes together in a probability problem?","back":"When the routes are mutually exclusive outcomes (e.g. \"Red-Red\" OR \"Green-Green\").","mcqOptions":[{"text":"When the routes are mutually exclusive outcomes (e.g. \"Red-Red\" OR \"Green-Green\").","correct":true,"explanation":"Mutually exclusive routes cannot happen simultaneously, so their probabilities are added when either route would satisfy the required outcome."},{"text":"When the routes are independent outcomes (e.g. \"Red-Red\" AND \"Green-Green\").","correct":false,"explanation":"Independent events do not automatically correspond to routes that should be added. Separate routes are added when they are mutually exclusive alternatives satisfying the required outcome."},{"text":"When the routes are consecutive outcomes (e.g. \"Red\" THEN \"Green\").","correct":false,"explanation":"Successive outcomes on one route are combined by multiplication, not addition. Addition applies to separate mutually exclusive routes that represent alternative successful outcomes."},{"text":"When the routes are equally likely outcomes (e.g. \"Red-Blue\" AND \"Blue-Red\").","correct":false,"explanation":"Equally likely routes need not be relevant alternatives, and equality is not the reason for adding. Add routes when they are mutually exclusive successful outcomes."}]},{"id":"0d4dc271-97d8-c8e4-0d82-4cff5eca2f26","cardType":"basic","points":2,"front":"How do you calculate the probability of picking \"one of each color\" from a bag of red and blue counters?","back":"Calculate the probability of Red-Blue and Blue-Red separately, then add them together.","mcqOptions":[{"text":"Calculate the probability of Red-Blue and Blue-Red separately, then add them together.","correct":true,"explanation":"One of each colour can occur in either order. These two mutually exclusive routes must be calculated separately and then added to obtain the total probability."},{"text":"Calculate the probability of Red-Blue and Blue-Red separately, then multiply them together.","correct":false,"explanation":"Multiplying the completed route probabilities would mean both mutually exclusive orders occur at once. Calculate Red-Blue and Blue-Red separately, then add their probabilities."},{"text":"Calculate the probability of Red-Red and Blue-Blue separately, then add them together.","correct":false,"explanation":"Red-Red and Blue-Blue represent two counters of the same colour, not one of each. Use the Red-Blue and Blue-Red routes and add them."},{"text":"Calculate the probability of Red-Blue and Blue-Red separately, then subtract them.","correct":false,"explanation":"Subtracting the two orders can cancel valid outcomes and does not represent either order occurring. Calculate both mixed-colour routes and add their probabilities."}]},{"id":"c5d42864-519f-0cf6-c6be-11aa0c53e6e4","cardType":"cloze","points":null,"front":"If the probability of picking a blue pen on Monday is 0.65, the probability of picking a black pen is [ ].","back":"If the probability of picking a blue pen on Monday is 0.65, the probability of picking a black pen is $1 - 0.65 = 0.35$.","mcqOptions":[{"text":"$$1 - 0.65 = 0.35$","correct":true,"explanation":"Blue and black are complementary outcomes, so their probabilities total 1. Subtracting 0.65 from 1 gives a black-pen probability of 0.35."},{"text":"$$0.65 - 1 = -0.35$","correct":false,"explanation":"This subtraction is reversed, producing a negative value, which cannot be a probability. The complement is found using $1 - 0.65 = 0.35$."},{"text":"$$1 + 0.65 = 1.65$","correct":false,"explanation":"Adding complementary probabilities is the wrong operation and gives a result above 1. The black probability is the complement, $1 - 0.65 = 0.35$."},{"text":"$$1 - 0.35 = 0.65$","correct":false,"explanation":"This treats the blue and black probabilities as equal. Since they are complementary, the black probability must be $1 - 0.65 = 0.35$."}]},{"id":"b47e6182-2075-5750-c7f6-f07dfffefbfc","cardType":"basic","points":2,"front":"A bag has 5 red and 3 blue counters. If a red is picked first and not replaced, what is $P(\\text{red second} \\mid \\text{red first})$?","back":"$$\\frac{4}{7}$","mcqOptions":[{"text":"$$\\frac{4}{7}$","correct":true,"explanation":"After the first red counter is removed, 4 red counters remain out of 7 counters altogether. Therefore, the conditional probability of another red is $\\frac{4}{7}$."},{"text":"$$\\frac{5}{7}$","correct":false,"explanation":"This keeps all 5 red counters in the numerator despite one already being removed. Only 4 of the 7 remaining counters are red, so the answer is $\\frac{4}{7}$."},{"text":"$$\\frac{4}{8}$","correct":false,"explanation":"This treats the first counter as though it were replaced, leaving 8 counters altogether. Without replacement, 4 red counters remain among 7, giving $\\frac{4}{7}$."},{"text":"$$\\frac{3}{7}$","correct":false,"explanation":"This uses the original number of blue counters as the numerator. After a red is removed, there are 4 red counters among 7 remaining, so the probability is $\\frac{4}{7}$."}]},{"id":"95910b3c-fc24-4dac-b2da-f5b5fd7a3ca6","cardType":"cloze","points":null,"front":"When picking two counters without replacement from a bag of 12, the denominator for the second branch will be [ ].","back":"When picking two counters without replacement from a bag of 12, the denominator for the second branch will be 11.","mcqOptions":[{"text":"11","correct":true,"explanation":"Without replacement, the first pick removes one counter from the original total of 12. Every second branch therefore has a denominator of $12-1=11$."},{"text":"12","correct":false,"explanation":"This incorrectly keeps the original total, as if the first counter were replaced. Without replacement, only 11 counters remain, so the second-branch denominator is 11."},{"text":"10","correct":false,"explanation":"This subtracts both picks before calculating the second pick. At the moment of the second pick, only one counter has been removed, so the denominator is 11."},{"text":"13","correct":false,"explanation":"This mistakenly increases the total after the first pick. Removing one counter from a bag of 12 leaves 11, so every second-branch denominator is 11."}]},{"id":"6a6ab753-d7f2-ed65-0624-91ca3746f95e","cardType":"basic","points":1,"front":"What does it mean for two routes to be \"mutually exclusive\"?","back":"They cannot both happen at the same time.","mcqOptions":[{"text":"They cannot both happen at the same time.","correct":true,"explanation":"Mutually exclusive routes have no shared outcomes, so one route occurring prevents the other from occurring simultaneously. Their intersection therefore has probability zero."},{"text":"They must both happen at the same time.","correct":false,"explanation":"This describes routes that always occur together, not mutually exclusive routes. Mutually exclusive routes share no outcomes and therefore cannot both happen at the same time."},{"text":"They do not affect each other when one happens.","correct":false,"explanation":"This confuses mutually exclusive events with independent events. Independence means one event does not affect the other, whereas mutually exclusive routes cannot both happen at the same time."},{"text":"They cover every possible outcome in the sample space.","correct":false,"explanation":"This describes exhaustive routes, which together cover the whole sample space. Mutually exclusive routes instead have no overlap, meaning they cannot both happen at the same time."}]},{"id":"b22bfc42-8603-ef0d-8d2e-3d95f9973adf","cardType":"basic","points":2,"front":"A bag has 7 red and 5 green counters. How do you set up the calculation for \"both counters are the same color\" (without replacement)?","back":"$$(\\frac{7}{12} \\times \\frac{6}{11}) + (\\frac{5}{12} \\times \\frac{4}{11})$","mcqOptions":[{"text":"$$(\\frac{7}{12} \\times \\frac{6}{11}) + (\\frac{5}{12} \\times \\frac{4}{11})$","correct":true,"explanation":"The two valid routes are red then red and green then green. Their branch probabilities are multiplied within each route, then the two mutually exclusive routes are added."},{"text":"$$(\\frac{7}{12} \\times \\frac{7}{11}) + (\\frac{5}{12} \\times \\frac{5}{11})$","correct":false,"explanation":"This fails to reduce each colour count after the first pick, effectively treating the counters as replaced. The second numerators should be 6 and 4, giving the correct sum."},{"text":"$$(\\frac{7}{12} \\times \\frac{5}{11}) + (\\frac{5}{12} \\times \\frac{7}{11})$","correct":false,"explanation":"These routes represent red then green and green then red, so they calculate different colours. For the same colour, use red-red and green-green instead."},{"text":"$$(\\frac{7}{12} \\times \\frac{6}{11}) \\times (\\frac{5}{12} \\times \\frac{4}{11})$","correct":false,"explanation":"This multiplies the red-red and green-green routes as though both routes must occur together. They are alternative outcomes, so their separate products must be added."}]},{"id":"c0d722e1-552d-6f6e-77ce-fcf442fce5dc","cardType":"basic","points":2,"front":"Why is the calculation for \"different colors\" usually more complex than for a specific sequence like \"Black then Red\"?","back":"It usually involves two different orders (BR and RB) that must be added.","mcqOptions":[{"text":"It usually involves two different orders (BR and RB) that must be added.","correct":true,"explanation":"Different colours can occur as Black then Red or Red then Black. Each order is a separate route whose probabilities are multiplied, and the two route probabilities are then added."},{"text":"It usually involves two different orders (BR and RB) that must be multiplied.","correct":false,"explanation":"Different colours do not require both orders to occur together. BR and RB are alternative routes, so their probabilities must be calculated separately and then added, not multiplied."},{"text":"It usually involves only one possible order (BR but not RB) in the calculation.","correct":false,"explanation":"This overlooks the reverse order, Red then Black. Different colours can occur through either BR or RB, so both route probabilities must be calculated and added."},{"text":"It usually involves replacing the first counter before the second counter is selected.","correct":false,"explanation":"Replacement affects the branch probabilities but does not explain the extra complexity. The key issue is that different colours includes both BR and RB, whose probabilities must be added."}]},{"id":"873fa758-3f9d-916c-2717-af840744986d","cardType":"cloze","points":null,"front":"In a Venn diagram, the region representing \"Tea and Coffee but not Juice\" excludes the [ ] where all three meet.","back":"In a Venn diagram, the region representing \"Tea and Coffee but not Juice\" excludes the triple overlap where all three meet.","mcqOptions":[{"text":"triple overlap","correct":true,"explanation":"The triple overlap contains people who are in Tea, Coffee and Juice. Excluding it leaves only those who are in both Tea and Coffee but not Juice."},{"text":"pairwise overlap","correct":false,"explanation":"The pairwise Tea and Coffee overlap includes the required region, so excluding all of it would remove the desired people. Only the triple overlap with Juice must be excluded."},{"text":"union","correct":false,"explanation":"The union contains everyone in at least one of the three sets, far beyond the region being excluded. Only the triple overlap belongs to all three sets."},{"text":"outside region","correct":false,"explanation":"The region outside the circles represents people in none of the sets, not people in all three. Tea and Coffee but not Juice excludes the triple overlap."}]},{"id":"8f1358f2-f808-fcdd-77da-53485e92a364","cardType":"basic","points":1,"front":"If 17 people like Tea and Coffee, and 12 of those also like Juice, how many people like Tea and Coffee only?","back":"$$17 - 12 = 5$","mcqOptions":[{"text":"$$17 - 12 = 5$","correct":true,"explanation":"The 17 people counted in Tea and Coffee include the 12 who also like Juice. Subtracting those 12 leaves 5 who like Tea and Coffee only."},{"text":"$$17 + 12 = 29$","correct":false,"explanation":"Adding double-counts the 12 people because they are already included among the 17 who like Tea and Coffee. Subtracting them gives $17-12=5$."},{"text":"$$17 - 5 = 12$","correct":false,"explanation":"This subtracts the number sought rather than the 12 people known to also like Juice. Tea and Coffee only is found directly using $17-12=5$."},{"text":"$$17 - 17 = 0$","correct":false,"explanation":"This assumes all 17 Tea-and-Coffee people also like Juice, although only 12 do. Removing those 12 from 17 leaves $17-12=5$ people."}]},{"id":"58433bac-7d9f-7767-a210-399a0a3c9e1a","cardType":"cloze","points":null,"front":"If there are 5 cards (1, 1, 2, 2, 4) and a 1 is picked first, the probability that the second card is greater than the first is [ ].","back":"If there are 5 cards (1, 1, 2, 2, 4) and a 1 is picked first, the probability that the second card is greater than the first is $\\frac{3}{4}$.","mcqOptions":[{"text":"$$\\frac{3}{4}$","correct":true,"explanation":"After one 1 is removed, four cards remain: 1, 2, 2 and 4. Three of these are greater than 1, so the probability is $\\frac{3}{4}$."},{"text":"$$\\frac{1}{2}$","correct":false,"explanation":"This counts only the two cards labelled 2 and overlooks that 4 is also greater than 1. Three of the four remaining cards are favourable, giving $\\frac{3}{4}$."},{"text":"$$\\frac{3}{5}$","correct":false,"explanation":"This retains the original denominator of 5 as if the first card were replaced. Only four cards remain, and three are greater than 1, giving $\\frac{3}{4}$."},{"text":"$$\\frac{1}{4}$","correct":false,"explanation":"This counts the one remaining card labelled 1, which is the only unfavourable outcome. The other three cards exceed 1, so the required probability is $\\frac{3}{4}$."}]},{"id":"e5201709-a3a0-2412-6f47-f35b4c92efe6","cardType":"basic","points":3,"front":"What is the 3-step strategy for calculating probability for \"second card greater than first\" using a tree?","back":"

Identify routes where second > first.
Multiply along those specific branches.
Add the resulting probabilities.

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

Identify routes where second > first.
Multiply along those specific branches.
Add the resulting probabilities.

","correct":true,"explanation":"Each valid route represents a mutually exclusive outcome. Multiplying gives the probability of each route, and adding those route probabilities gives the overall probability that the second card is greater."},{"text":"

Identify routes where second > first.
Add along those specific branches.
Multiply the resulting probabilities.

","correct":false,"explanation":"This reverses the tree rules. Probabilities are multiplied along each valid route, then the probabilities of the separate valid routes are added together."},{"text":"

Identify routes where second < first.
Multiply along those specific branches.
Add the resulting probabilities.

","correct":false,"explanation":"This selects the opposite event, where the second card is smaller. The correct method selects routes where second > first, multiplies along them, then adds their probabilities."},{"text":"

Identify routes where second > first.
Multiply all branches in the tree.
Add the resulting probabilities.

","correct":false,"explanation":"Including every branch counts outcomes that do not satisfy the condition. Only routes where second > first should be multiplied individually before their probabilities are added."}]},{"id":"6578a929-1f2d-0e03-a4ef-a709fa6b568b","cardType":"basic","points":2,"front":"In a Venn-style probability question, how do you handle picking \"two people\" from a group?","back":"Treat it as conditional probability without replacement; the total group size decreases for the second pick.","mcqOptions":[{"text":"Treat it as conditional probability without replacement; the total group size decreases for the second pick.","correct":true,"explanation":"Once the first person has been chosen, that person is unavailable for the second choice. Therefore, the second probability is conditional and uses a total reduced by one."},{"text":"Treat it as independent probability with replacement; the total group size stays the same for the second pick.","correct":false,"explanation":"This incorrectly assumes the first person is returned to the group. People are normally chosen without replacement, so the second probability depends on the first pick and has a smaller total."},{"text":"Treat it as independent probability without replacement; the total group size decreases for the second pick.","correct":false,"explanation":"Although the group size decreases, the events are not independent because the first selection changes the second selection’s probabilities. It is conditional probability without replacement."},{"text":"Treat it as conditional probability with replacement; the total group size increases for the second pick.","correct":false,"explanation":"Replacement would keep the group size unchanged, not increase it, and is not normally used when choosing two different people. The correct model is conditional probability without replacement."}]},{"id":"036b43b5-ef82-795c-d48f-bad57c41e3b0","cardType":"cloze","points":null,"front":"To find the probability of a complement (the event not happening), you subtract the probability of the event from [ ].","back":"To find the probability of a complement (the event not happening), you subtract the probability of the event from 1.","mcqOptions":[{"text":"1","correct":true,"explanation":"The probabilities of an event and its complement sum to 1. Therefore, the probability that the event does not happen is $1-P(A)$."},{"text":"0","correct":false,"explanation":"Subtracting from 0 confuses the minimum possible probability with the total probability. An event and its complement exhaust all outcomes, so their probabilities add to 1."},{"text":"$$-1$","correct":false,"explanation":"Probabilities cannot be negative, so $-1$ cannot represent the total probability. Complement probabilities are found by subtracting the event’s probability from 1."},{"text":"2","correct":false,"explanation":"A total of 2 lies outside the probability scale, which runs from 0 to 1. The event and its complement together have probability 1."}]},{"id":"b6898156-8a3e-e3c1-0a4d-fe57e0b8a3e7","cardType":"basic","points":2,"front":"If 29 students like coffee out of 40 total, what is the probability two chosen students both like coffee?","back":"$$\\frac{29}{40} \\times \\frac{28}{39}$","mcqOptions":[{"text":"$$\\frac{29}{40} \\times \\frac{28}{39}$","correct":true,"explanation":"The first coffee-liker is chosen with probability $\\frac{29}{40}$. Without replacement, 28 coffee-likers remain among 39 students for the second choice."},{"text":"$$\\frac{29}{40} \\times \\frac{29}{40}$","correct":false,"explanation":"This treats the selections as if the first student were replaced. Without replacement, both the number of coffee-likers and the total number of students decrease by one."},{"text":"$$\\frac{29}{40} \\times \\frac{28}{40}$","correct":false,"explanation":"This reduces the number of coffee-likers but forgets to reduce the total group size. After one student is chosen, only 39 students remain."},{"text":"$$\\frac{29}{40} \\times \\frac{29}{39}$","correct":false,"explanation":"This reduces the total group size but not the number of coffee-likers. After choosing a student who likes coffee, only 28 such students remain among the other 39."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Unit 1"},"heading":"Conditional Probability","title":"Conditional Probability Flashcards","topbar":{"backHref":"/gcse/maths/wjec/flashcards"},"topicSidebar":"$L39"}]}],"topicId":"b453f9cc-2d2d-12ad-8c62-9f0de647ce6e","walled":"$L3a"}],"$L3b","$L3c"]