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together.","correct":true,"explanation":"Following one specific path means both sequential outcomes must occur. The multiplication rule therefore combines their individual probabilities to find the probability of the complete path."},{"text":"Add the probabilities of the individual outcomes together.","correct":false,"explanation":"Adding probabilities is used for mutually exclusive alternative outcomes, not successive events on one path. For two sequential outcomes, multiply the individual probabilities together."},{"text":"Subtract the probabilities of the individual outcomes.","correct":false,"explanation":"Subtraction may be used when finding a complement, but it does not combine sequential events. Along a specific path, multiply the probabilities of the individual outcomes together."},{"text":"Divide the probabilities of the individual outcomes.","correct":false,"explanation":"Division can arise in conditional probability calculations, but it is not the path rule. The probability of successive outcomes along one path is found by multiplying."}]},{"id":"54dc1ad4-d6b7-4df7-a5e8-a67d60bbeda9","cardType":"cloze","points":null,"front":"In without replacement questions, the denominator for the second pick is usually [ ] than the first denominator.","back":"In without replacement questions, the denominator for the second pick is usually one less than the first denominator.","mcqOptions":[{"text":"one less","correct":true,"explanation":"Without replacement, the first selected item is not returned, so the total number of available items decreases by one before the second pick."},{"text":"one greater","correct":false,"explanation":"This incorrectly treats the first pick as adding an item to the bag. Without replacement, one item is removed, so the second denominator is one less."},{"text":"the same","correct":false,"explanation":"Keeping the denominator unchanged is appropriate with replacement. Without replacement, the first item remains removed, making the denominator for the second pick one less."},{"text":"two less","correct":false,"explanation":"Only one item has been selected before the second pick, so reducing the denominator by two is an off-by-one error. It should be one less."}]},{"id":"f37306ba-6689-4b1b-ab4b-99b01bd92e43","cardType":"basic","points":1,"front":"A bag contains $x$ blue counters and 5 red counters. What is the total number of counters?","back":"$$x + 5$","mcqOptions":[{"text":"$$x + 5$","correct":true,"explanation":"The total is found by adding the numbers of each colour. There are $x$ blue counters and 5 red counters, giving $x + 5$ altogether."},{"text":"$$x - 5$","correct":false,"explanation":"Subtracting would find a difference between the numbers of counters, not the total. Adding the $x$ blue and 5 red counters gives $x + 5$."},{"text":"$$5x$","correct":false,"explanation":"Multiplication would mean there are five groups of $x$ counters, which is not stated. The two separate quantities must be added, giving $x + 5$."},{"text":"$$rac{x}{5}$","correct":false,"explanation":"Dividing compares the number of blue counters with the number of red counters rather than finding the total. The total number is $x + 5$."}]},{"id":"1ea29bae-f9d9-4a58-999e-5f1909038f61","cardType":"cloze","points":null,"front":"In with replacement questions, the probabilities for the second pick [ ] as the first pick.","back":"In with replacement questions, the probabilities for the second pick stay the same as the first pick.","mcqOptions":[{"text":"stay the same","correct":true,"explanation":"With replacement, the selected item is returned before the second pick. The bag’s composition is restored, so all probabilities remain unchanged."},{"text":"decrease by one","correct":false,"explanation":"This confuses the number of items with the probability and ignores replacement. Returning the selected item restores the original composition, so the probabilities stay the same."},{"text":"always increase","correct":false,"explanation":"Replacement does not increase the number or proportion of any type of item. It restores the original contents, so the second-pick probabilities stay the same."},{"text":"always decrease","correct":false,"explanation":"Probabilities decrease only in some without-replacement situations, depending on the first outcome. With replacement, the original contents are restored and the probabilities stay the same."}]},{"id":"65fbbf5e-bd38-4c15-9065-6680602505e0","cardType":"basic","points":2,"front":"How should you represent the number of red and blue counters if their ratio is 3:1?","back":"

Red: $3k$
Blue: $k$

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

Red: $3k$
Blue: $k$

","correct":true,"explanation":"A ratio of 3:1 means the number of red counters is three equal parts and the number of blue counters is one part, represented by $3k$ and $k$."},{"text":"

Red: $k$
Blue: $3k$

","correct":false,"explanation":"This reverses the order of the ratio and represents red to blue as 1:3. For 3:1, red must be $3k$ and blue must be $k$."},{"text":"

Red: $3k$
Blue: $2k$

","correct":false,"explanation":"These expressions give a red-to-blue ratio of 3:2 rather than 3:1. The blue quantity must be one equal part, so it should be $k$."},{"text":"

Red: $4k$
Blue: $k$

","correct":false,"explanation":"These expressions represent a ratio of 4:1, adding an extra part to the red quantity. A 3:1 ratio is represented by $3k$ and $k$."}]},{"id":"934f0c17-3b53-4ee8-9d83-b1744c2cfbf3","cardType":"basic","points":1,"front":"A bag has $x$ blue counters. If one blue counter is removed and not replaced, what is the new number of blue counters?","back":"$$x - 1$","mcqOptions":[{"text":"$$x - 1$","correct":true,"explanation":"Removing one of the $x$ blue counters reduces the number of blue counters by exactly one. Because it is not replaced, the new number is $x - 1$."},{"text":"$$x + 1$","correct":false,"explanation":"Adding one would represent a blue counter being put into the bag rather than removed. Removing one without replacement leaves $x - 1$ blue counters."},{"text":"$$x$","correct":false,"explanation":"The number would remain $x$ only if the removed blue counter were replaced. Since it is not returned, the new number is $x - 1$."},{"text":"$$x - 2$","correct":false,"explanation":"Subtracting two assumes that two blue counters were removed. Only one blue counter is removed and not replaced, so the new number is $x - 1$."}]},{"id":"0067c1ec-ba8b-4112-a358-72bafaa69573","cardType":"cloze","points":null,"front":"A probability equation is formed by setting an [ ] equal to a given probability value.","back":"A probability equation is formed by setting an algebraic probability equal to a given probability value.","mcqOptions":[{"text":"algebraic probability","correct":true,"explanation":"An algebraic probability expresses the probability using an unknown, often derived from numbers of outcomes. Equating it to the stated probability creates an equation that can be solved."},{"text":"experimental frequency","correct":false,"explanation":"An experimental frequency is a count from observed trials, not the unknown probability expression used to form the equation. The required quantity is an algebraic probability."},{"text":"sample space","correct":false,"explanation":"A sample space lists possible outcomes rather than providing an expression to equate to a probability value. A probability equation begins with an algebraic probability."},{"text":"outcome total","correct":false,"explanation":"The total number of outcomes may form part of a probability fraction, but it is not the complete expression being equated. The equation uses an algebraic probability."}]},{"id":"e5f28e9b-a0cc-4c83-98c9-0dc483281fda","cardType":"cloze","points":null,"front":"A probability equation is formed by setting an algebraic probability equal to a [ ] value.","back":"A probability equation is formed by setting an algebraic probability equal to a given probability value.","mcqOptions":[{"text":"given probability","correct":true,"explanation":"The algebraic expression models the required event’s probability. Setting it equal to the probability supplied in the question produces an equation that can be solved for the unknown."},{"text":"total frequency","correct":false,"explanation":"A total frequency is a number of observations, not necessarily a probability between 0 and 1. The algebraic probability must equal the given probability value."},{"text":"sample-space size","correct":false,"explanation":"The sample-space size counts possible outcomes and may be used as a denominator, but it is not the stated probability value. The equation uses the given probability."},{"text":"expected frequency","correct":false,"explanation":"Expected frequency is found by multiplying probability by the number of trials, so it is not itself the required probability value. The expression is set equal to the given probability."}]},{"id":"c4c5170a-c089-4e5d-a089-6565db54a2db","cardType":"basic","points":1,"front":"If the ratio of red to blue counters is 2:1, what is the algebraic expression for the total number of counters?","back":"$$3k$ (where red is $2k$ and blue is $k$).","mcqOptions":[{"text":"$$3k$ (where red is $2k$ and blue is $k$).","correct":true,"explanation":"The ratio 2:1 has 3 equal parts altogether, so the counters can be represented by $2k$ red and $k$ blue, giving a total of $3k$."},{"text":"$$2k$ (where red is $k$ and blue is $k$).","correct":false,"explanation":"This treats the two ratio parts as equal and therefore loses the 2:1 relationship. The total must contain three parts, so the correct expression is $3k$."},{"text":"$$k$ (where red is $2k$ and blue is $k$).","correct":false,"explanation":"This gives only the blue portion rather than adding both colours. Since red contributes $2k$ and blue contributes $k$, the total is $3k$."},{"text":"$$4k$ (where red is $2k$ and blue is $2k$).","correct":false,"explanation":"This changes the ratio to 1:1 by making both colours $2k$. The ratio 2:1 requires $2k$ red and $k$ blue, producing a total of $3k$."}]},{"id":"71863e6d-5db2-4c6a-8202-8ec55918216e","cardType":"basic","points":2,"front":"When solving $\\left(\\frac{4}{y+4}\\right)^2 = \\frac{4}{25}$, why is only the positive square root used?","back":"

Probabilities cannot be negative.
The number of counters must be positive.

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

Probabilities cannot be negative.
The number of counters must be positive.

","correct":true,"explanation":"A probability cannot be negative, and a bag containing counters must have a positive number of counters. Therefore, only the positive square-root result is valid."},{"text":"

Probabilities must be negative.
The number of counters may be zero.

","correct":false,"explanation":"Probabilities are never negative, and a bag described as containing counters cannot have zero counters. The valid result is selected because probabilities are non-negative and the counter count is positive."},{"text":"

Probabilities must be greater than one.
The number of counters must be even.

","correct":false,"explanation":"Probabilities range from 0 to 1, not above 1, and there is no requirement that the number of counters be even. The positive root is chosen for the stated probability and count restrictions."},{"text":"

Probabilities can be negative.
The number of counters must be positive.

","correct":false,"explanation":"Although a counter total must be positive, probabilities cannot be negative. The negative square-root result must therefore be rejected, leaving only the positive root."}]},{"id":"db01d8e7-8dfb-4770-b0c1-4c1329b04738","cardType":"cloze","points":null,"front":"When a probability equation results in a quadratic, you must [ ] any solutions that are negative or non-integers.","back":"When a probability equation results in a quadratic, you must reject any solutions that are negative or non-integers.","mcqOptions":[{"text":"reject","correct":true,"explanation":"A quadratic may produce algebraic solutions that are negative or non-integers, but these cannot represent valid counter counts. Such solutions must be rejected."},{"text":"accept","correct":false,"explanation":"Accepting every quadratic solution could include negative or fractional counter counts, which are impossible in a probability model. These invalid solutions must be rejected."},{"text":"round","correct":false,"explanation":"Rounding an invalid solution can change the equation and create a value that was not actually obtained. Negative or non-integer solutions should be rejected rather than rounded."},{"text":"double","correct":false,"explanation":"Changing the size of a solution does not test whether it represents a valid number of counters. Negative or non-integer solutions must be rejected altogether."}]},{"id":"4b4fe540-1620-4e29-9f7d-e886887ab17c","cardType":"cloze","points":null,"front":"When a probability equation results in a quadratic, you must reject any solutions that are [ ] or non-integers.","back":"When a probability equation results in a quadratic, you must reject any solutions that are negative or non-integers.","mcqOptions":[{"text":"negative","correct":true,"explanation":"A number of counters cannot be negative, so any negative quadratic solution is impossible in this context. Non-integer solutions must also be rejected because counters are counted in whole numbers."},{"text":"positive","correct":false,"explanation":"Positive solutions are usually the possible ones for a counter count, so they should not automatically be rejected. The invalid category here is negative solutions, along with non-integers."},{"text":"zero","correct":false,"explanation":"Zero is not generally the category that must be rejected; it can sometimes be a valid boundary value depending on the model. Negative values are impossible for numbers of counters."},{"text":"fractional","correct":false,"explanation":"Fractional values are already covered by non-integers, so this does not identify the separate impossible category. The missing category is negative solutions, which cannot represent counters."}]},{"id":"d87f2866-608e-40e1-a63a-1a9c69dd8b55","cardType":"basic","points":1,"front":"A bag has $n$ counters; 6 are red and the rest are blue. What is the expression for the number of blue counters?","back":"$$n - 6$","mcqOptions":[{"text":"$$n - 6$","correct":true,"explanation":"The total number of counters is $n$, and 6 of them are red. Subtracting the red counters from the total leaves $n - 6$ blue counters."},{"text":"$$n + 6$","correct":false,"explanation":"Adding the red counters to the total double-counts them rather than removing them. The blue counters are what remains after subtracting 6 from $n$, giving $n - 6$."},{"text":"$$6 - n$","correct":false,"explanation":"Reversing the subtraction usually gives a negative value when the total exceeds 6, which cannot represent blue counters. The correct remainder is found with $n - 6$."},{"text":"$$\\frac{n}{6}$","correct":false,"explanation":"Dividing by 6 does not separate the red counters from the total; it changes the scale instead. Removing the 6 red counters requires subtraction, so the expression is $n - 6$."}]},{"id":"52d00342-23b2-4896-823f-d1b2beb18673","cardType":"basic","points":2,"front":"The ratio of red:blue is 2:1. Two counters are picked without replacement. Write the expression for $P(\\text{red then red})$.","back":"$$$\\frac{2k}{3k} \\times \\frac{2k - 1}{3k - 1}$$","mcqOptions":[{"text":"$$$\\frac{2k}{3k} \\times \\frac{2k - 1}{3k - 1}$$","correct":true,"explanation":"Initially, the probability of red is $\\frac{2k}{3k}$. After one red is removed, $2k-1$ red counters remain out of $3k-1$ total counters, giving the product shown."},{"text":"$$$\\frac{2k}{3k} \\times \\frac{2k}{3k - 1}$$","correct":false,"explanation":"This forgets that the first red counter has been removed, so the second red count should decrease as well as the total. The correct second factor is $\\frac{2k-1}{3k-1}$."},{"text":"$$$\\frac{2k}{3k} \\times \\frac{2k - 1}{3k}$$","correct":false,"explanation":"Without replacement, the total number of counters decreases after the first pick, so the second denominator cannot remain $3k$. It must become $3k-1$, giving the stated product."},{"text":"$$$\\frac{k}{3k} \\times \\frac{k - 1}{3k - 1}$$","correct":false,"explanation":"This uses the blue count $k$ instead of the red count $2k$, calculating the probability of blue then blue. Red then red requires $2k$ red counters initially."}]},{"id":"7272f4da-7365-4fe9-b421-56c37ab5a768","cardType":"basic","points":1,"front":"Why might a solution like $n = 4$ be rejected if a bag is known to contain 6 red counters?","back":"The total number of counters ($n$) cannot be less than the number of red counters already in the bag.","mcqOptions":[{"text":"The total number of counters ($n$) cannot be less than the number of red counters already in the bag.","correct":true,"explanation":"The total includes all 6 red counters, so it must be at least 6. Since 4 is less than 6, it cannot be a possible value for $n$."},{"text":"The total number of counters ($n$) cannot be greater than the number of red counters already in the bag.","correct":false,"explanation":"A bag may contain blue counters as well as the 6 red counters, so its total can be greater than 6. The total cannot be less than the number of red counters."},{"text":"The total number of counters ($n$) must be equal to the number of red counters already in the bag.","correct":false,"explanation":"This would incorrectly assume that the bag contains no blue counters. The total may exceed 6, but it cannot be below 6 because all red counters are included in the total."},{"text":"The total number of counters ($n$) cannot be less than or equal to the number of red counters already in the bag.","correct":false,"explanation":"A total of exactly 6 is possible if every counter is red, so equality is not forbidden. Only totals below 6 are impossible because the bag already contains 6 red counters."}]},{"id":"d64380ed-c74e-4d4c-808b-aba88ff7b2f8","cardType":"cloze","points":null,"front":"To solve $\\frac{x(x - 1)}{(x + 3)(x + 2)} = \\frac{5}{14}$, the first step is to [ ] the fractions.","back":"To solve $\\frac{x(x - 1)}{(x + 3)(x + 2)} = \\frac{5}{14}$, the first step is to cross-multiply the fractions.","mcqOptions":[{"text":"cross-multiply","correct":true,"explanation":"Cross-multiplication removes the two denominators by multiplying opposite numerator and denominator terms, producing an equation that can then be expanded and rearranged into a quadratic."},{"text":"factor","correct":false,"explanation":"Factoring is normally used after the equation has been rearranged into a suitable polynomial form. The first step here is to remove the fractions by cross-multiplying."},{"text":"simplify","correct":false,"explanation":"The fractions cannot be cancelled directly because their factors are not common across the equality. Cross-multiplication is the initial operation used to eliminate the denominators."},{"text":"add","correct":false,"explanation":"Adding the fractions would not correctly preserve the equation or remove the denominators. Multiplying opposite sides across the equality is the required first step."}]},{"id":"a6f58302-8552-4059-82fb-4e586880bdeb","cardType":"basic","points":2,"front":"If the first pick has a probability of $\\frac{x}{x+3}$, what is the probability of picking the same item again without replacement?","back":"$$$\\frac{x - 1}{x + 2}$$","mcqOptions":[{"text":"$$$\\frac{x - 1}{x + 2}$$","correct":true,"explanation":"After the first item is selected, one matching item and one total item have been removed. The second probability is therefore $(x-1)/(x+2)$."},{"text":"$$$\\frac{x}{x + 2}$$","correct":false,"explanation":"This reduces the total number of items but not the number of matching items. Without replacement, both quantities decrease by one, giving $(x-1)/(x+2)$."},{"text":"$$$\\frac{x - 1}{x + 3}$$","correct":false,"explanation":"This reduces the number of matching items but leaves the original total unchanged. One item has also left the bag, so the denominator must become $x+2$."},{"text":"$$$\\frac{x}{x + 3}$$","correct":false,"explanation":"This simply repeats the first-pick probability and treats the selection as though replacement occurred. Without replacement, the second probability is $(x-1)/(x+2)$."}]},{"id":"874724b4-5195-4851-ac0f-96673b9d4d1f","cardType":"cloze","points":null,"front":"In a \"show that\" question, you must begin with the [ ] and expand it until it matches the required quadratic equation.","back":"In a \"show that\" question, you must begin with the probability statement and expand it until it matches the required quadratic equation.","mcqOptions":[{"text":"probability statement","correct":true,"explanation":"The probability statement expresses the given event algebraically. Expanding and rearranging this statement then produces the quadratic equation that the question asks you to show."},{"text":"quadratic equation","correct":false,"explanation":"The quadratic equation is the result that must be reached, not the starting point. Begin with the probability statement, then expand and rearrange it."},{"text":"final answer","correct":false,"explanation":"The final answer cannot serve as the derivation in a “show that” question. The working must start from the probability statement and lead to the required equation."},{"text":"given information","correct":false,"explanation":"The given information helps construct the model, but the algebraic demonstration starts from the resulting probability statement, which can then be expanded and rearranged."}]},{"id":"2858242f-51d3-44dd-9412-2d921f276f4e","cardType":"cloze","points":null,"front":"In a \"show that\" question, you must begin with the probability statement and expand it until it matches the required [ ].","back":"In a \"show that\" question, you must begin with the probability statement and expand it until it matches the required quadratic equation.","mcqOptions":[{"text":"quadratic equation","correct":true,"explanation":"Expanding products of algebraic probabilities commonly creates a squared term. Rearranging the resulting probability statement therefore produces the required quadratic equation."},{"text":"linear equation","correct":false,"explanation":"A linear equation has no squared variable term, whereas multiplying algebraic probabilities generally produces one. The expression must be rearranged to match the required quadratic equation."},{"text":"probability tree","correct":false,"explanation":"A probability tree organises possible outcomes but is not the algebraic result requested. The probability statement must be expanded until it matches the required quadratic equation."},{"text":"numerical answer","correct":false,"explanation":"A numerical answer may follow after solving, but the “show that” stage requires a specific algebraic form. Here, the expanded statement must match the quadratic equation."}]},{"id":"60f8f16c-3cbc-46fc-8ca7-961cf07c26c6","cardType":"basic","points":2,"front":"A bag contains 5 red and 3 blue counters. Find $P(\\text{blue then blue})$ without replacement.","back":"$$$\\frac{3}{8} \\times \\frac{2}{7} = \\frac{6}{56}$$ (or $\\frac{3}{28}$)","mcqOptions":[{"text":"$$$\\frac{3}{8} \\times \\frac{2}{7} = \\frac{6}{56}$$ (or $\\frac{3}{28}$)","correct":true,"explanation":"The first blue probability is $3/8$. Without replacement, two blue counters remain among seven counters, so multiplication gives $6/56=3/28$."},{"text":"$$$\\frac{3}{8} \\times \\frac{3}{8} = \\frac{9}{64}$$ (or $\\frac{9}{64}$)","correct":false,"explanation":"Using $3/8$ for both picks incorrectly assumes replacement. Without replacement, the second probability is $2/7$, so the required result is $6/56=3/28$."},{"text":"$$$\\frac{3}{8} \\times \\frac{2}{8} = \\frac{6}{64}$$ (or $\\frac{3}{32}$)","correct":false,"explanation":"The blue numerator has decreased correctly, but the denominator must also decrease because only seven counters remain. The correct product is $3/8\\times2/7=3/28$."},{"text":"$$$\\frac{3}{8} \\times \\frac{3}{7} = \\frac{9}{56}$$ (or $\\frac{9}{56}$)","correct":false,"explanation":"The denominator reflects the removal, but the numerator incorrectly leaves three blue counters available. Only two remain, giving $3/8\\times2/7=3/28$."}]},{"id":"19dc7fe0-c0d7-4063-b667-55bf702849eb","cardType":"basic","points":1,"front":"Simplify the algebraic fraction $\\frac{2k}{3k}$ before solving a probability equation.","back":"$$$\\frac{2}{3}$$","mcqOptions":[{"text":"$$$\\frac{2}{3}$$","correct":true,"explanation":"The common non-zero factor $k$ cancels from the numerator and denominator: $2k/(3k)=2/3$. This simplifies the later probability equation."},{"text":"$$$\\frac{2k}{3}$$","correct":false,"explanation":"This cancels $k$ from only the denominator rather than from both parts of the fraction. Cancelling the common factor completely gives $2/3$."},{"text":"$$$\\frac{2}{3k}$$","correct":false,"explanation":"This removes $k$ from the numerator but incorrectly retains it in the denominator. The same factor must cancel from both, leaving $2/3$."},{"text":"$$$\\frac{k}{3}$$","correct":false,"explanation":"Cancelling the coefficients instead of the shared factor $k$ changes the fraction’s value. Only $k$ cancels, so the simplified fraction is $2/3$."}]},{"id":"158dea65-389f-4ab8-ac98-d241fb8dd657","cardType":"basic","points":3,"front":"What three algebraic values should you define first for a ratio question (e.g., Red:Blue is 3:1)?","back":"

Red counters ($3k$)
Blue counters ($k$)
Total counters ($4k$)

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

Red counters ($3k$)
Blue counters ($k$)
Total counters ($4k$)

","correct":true,"explanation":"The ratio $3:1$ gives $3k$ red counters and $k$ blue counters. Adding both parts gives the total number of counters as $4k$."},{"text":"

Red counters ($3k$)
Blue counters ($k$)
Total counters ($2k$)

","correct":false,"explanation":"The red and blue expressions match the ratio, but the total is found by addition, not subtraction. It must be $3k+k=4k$."},{"text":"

Red counters ($k$)
Blue counters ($3k$)
Total counters ($4k$)

","correct":false,"explanation":"This reverses the two parts of the stated Red:Blue ratio. A ratio of $3:1$ requires $3k$ red, $k$ blue and $4k$ altogether."},{"text":"

Red counters ($k + 3$)
Blue counters ($k + 1$)
Total counters ($2k + 4$)

","correct":false,"explanation":"The ratio numbers represent multiples of the same scale factor, not fixed counter values added to $k$. The correct expressions are $3k$, $k$ and $4k$."}]},{"id":"34c843b0-6a1e-4b9e-ace1-cf1635cc3d61","cardType":"basic","points":2,"front":"A bag has 4 red and $y$ blue counters. Two are picked with replacement. Write the expression for $P(\\text{red then red})$.","back":"$$$\\left(\\frac{4}{y + 4}\\right)^2$$","mcqOptions":[{"text":"$$$\\left(\\frac{4}{y + 4}\\right)^2$$","correct":true,"explanation":"There are four red counters among $y+4$ counters. Replacement leaves the probability unchanged for the second pick, so the probability is the square of $4/(y+4)$."},{"text":"$$$\\frac{4}{y + 4} \\times \\frac{3}{y + 3}$$","correct":false,"explanation":"This uses the without-replacement probabilities by reducing both counts after the first pick. With replacement, both red-pick probabilities remain $4/(y+4)$."},{"text":"$$$\\left(\\frac{y}{y + 4}\\right)^2$$","correct":false,"explanation":"This calculates the probability of two blue counters rather than two red counters. For red then red, the repeated probability is $4/(y+4)$."},{"text":"$$$\\frac{16}{y + 4}$$","correct":false,"explanation":"Squaring a fraction requires squaring both its numerator and denominator. The denominator must be $(y+4)^2$, giving $(4/(y+4))^2$."}]}]}],"examCountdown":null,"heading":"Probability Equation Questions","title":"Probability Equation Questions Flashcards","topbar":{"backHref":"/igcse/maths/oxford-aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"e5dcfad4-43b9-43cf-bba1-db68d9ac5316","walled":"$L39"}],"$L3a","$L3b"]