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This page sets out the method for probability equation questions in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Probability equation questions topic page.
When you see 'the probability that both … is . Find ', do this.
When you have an equation like , do this.
When part (a) says 'show that ' and part (b) says 'hence find ', do this.
The second draw has one fewer counter. The denominator changes.
P(both red) is the product of the two fractions along that path.
If that product equals a fraction, multiply out. You get n(n-1) equal to a number.
n(n-1) = N becomes n^2 - n - N = 0.
A negative root, or a root smaller than the number of red counters, cannot be the bag.
Read the probability off the question. Do not invent a different event.
Name the two fractions before you look at the given probability. There are n counters and r are red, so the first draw is r/n. Without replacement, one red counter has gone and one counter has gone, so the second draw is (r - 1)/(n - 1). Both draws have to be red. They sit on one route, so the probability is the product. Set that product equal to the fraction in the question. Do not invent a different event, such as one red or at least one red.
A small check shows the shape. Suppose 5 counters are red and the probability both are red is 2/3. Then 5/n × 4/(n - 1) = 2/3. The 4 is 5 - 1, not another 5. If the question had said with replacement, the second fraction would be 5/n again and the equation would not be quadratic in this way. Read the sentence about replacement before you write the second denominator.
Method: 1. Write r/n for the first red counter. 2. Write (r - 1)/(n - 1) for the second. 3. Multiply them, because both draws happen. 4. Set the product equal to the given fraction.
The product is a single fraction (r(r - 1))/(n(n - 1)). If that equals a/b, cross-multiply: r(r - 1) × b = a × n(n - 1). Then divide by a. What is left is n(n - 1) = N for a known integer N. With r = 9 and the probability 4/5, the red product is 9 × 8 = 72, and 72 × 5 ÷ 4 = 90. So n(n - 1) = 90. The n has not been found yet. You have found the product of the two consecutive whole numbers that surround it.
Dividing the wrong way is the usual slip. The correct value is 72 × 5 ÷ 4 = 90. Dividing by 5 and multiplying by 4 instead gives 57.6. Stopping after 72 ÷ 4 = 18 forgets to use the 5. Keep the given probability as one fraction until the cross-multiply is done. Then N should be an integer you recognise as a product of two consecutive whole numbers, such as 10 × 9 = 90 or 8 × 7 = 56.
Method: 1. Multiply the two red numerators. 2. Cross-multiply by the denominator of the given probability. 3. Divide by the numerator of the given probability. 4. Write n(n - 1) = N.
Expand the product. n(n - 1) = N becomes n^2 - n = N, then n^2 - n - N = 0. Compare with an^2 + bn + c = 0. Here a = 1, b = -1 and c = -N. The discriminant is b^2 - 4ac = (-1)^2 - 4(1)(-N) = 1 + 4N. That is the honest source of the 1 + 4N under the square root. It is not an extra guess. For N = 90, the discriminant is 1 + 360 = 361, and √(361) = 19.
The quadratic formula then says n = (1 ± 19)/2. The two roots are 20/2 = 10 and (-18)/2 = -9. You can also spot 10 × 9 = 90 once N is known, which finds the positive root without writing the formula. Either route is valid if the working is shown. The formula is the one to use when the consecutive pair is not obvious. For N = 210, the discriminant is 1 + 840 = 841 = 29^2, and the positive root is (1 + 29)/2 = 15.
Method: 1. Rearrange to n^2 - n - N = 0. 2. Write the discriminant as 1 + 4N. 3. Take the square root. 4. Apply n = dfrac1 ± √(1 + 4N)2.
A quadratic of this shape always has one positive root and one negative root, because the square root of 1 + 4N is bigger than 1. The negative root satisfies the algebra. It does not satisfy the story. You cannot have -9 counters in a bag. Reject it in a sentence, not by rubbing it out with no reason. The answer line is the positive root alone.
Check the positive root in the original fractions. For n = 10 and 9 red counters, 9/10 × 8/9 = 4/5, which is the probability you were given. Also check n > r. A root smaller than the number of red counters cannot hold them. If both checks pass, the positive root is the number of counters. Put that number, and not the rejected root, on the answer line.
Method: 1. Calculate both roots, or spot the consecutive pair and the negative partner. 2. Reject the negative root because a count cannot be negative. 3. Confirm the positive answer is larger than the red count in the bag. 4. Substitute it back into the two fractions.
Many exam questions ask for the probability that two counters are "the same colour" or "different colours". Same colour has two cases that must be added: both red, or both blue. Different colours also has two cases: red then blue, or blue then red. These two routes have equal probabilities, so you can work out one and double it. Forgetting the second case is the commonest way to lose marks.
Each case is a product of two fractions, and without replacement the second fraction has both a smaller numerator and a smaller denominator. When the counts are unknown, write them as algebra on the fractions. The result is a single fraction in x that is set equal to the probability you were given, and the equation to solve comes from cross-multiplying.
Method: 1. List the routes that give the event, such as RR and BB for same colour. 2. Write each route as two fractions, reducing the total by one for the second draw. 3. Multiply along each route, then add the routes together. 4. Put the sum equal to the given probability, ready to cross-multiply.
Tip: For two different colours, work out one route and double it. It halves the algebra.
An exam question rarely says "there are n counters". It says things like "there are 3 more blue counters than green counters". Let the unknown count be x green. Then there are x + 3 blue, and the total is 2x + 3. After one counter is taken, the total drops to 2x + 2. Write each count as a bracket so the algebra stays tidy.
The second draw is where most errors happen. If the first counter is green, the count of green falls by one but the blue stays the same. The total falls by one in every case. So the green-then-green route is x/(2x + 3) × (x - 1)/(2x + 2), and the blue-then-blue route is (x + 3)/(2x + 3) × (x + 2)/(2x + 2). Both share the same denominator, so they can be added with one numerator.
Method: 1. Let one colour be x and write the other colours in terms of x. 2. Write the total as an expression, then the total after one counter has gone. 3. Reduce only the count of the colour that was taken on the second draw. 4. Combine the routes over the shared denominator and expand the brackets.
Another common set-up changes the bag. A probability is given, some counters are added or removed, and a new probability is given. Let r be red and g be green. Each probability gives an equation after cross-multiplying. For example, g/(r + g) = 2/5 becomes 5g = 2r + 2g, which simplifies to 3g = 2r.
The second probability gives a second equation in the same two letters, and you solve the pair by substitution or elimination. Check the final counts by putting them back into both fractions. The answers must be whole numbers, because you cannot have half a counter. If a fraction answer appears, an arithmetic slip has happened, so check the cross-multiplying before moving on. Writing each equation on its own line makes the slip easy to find.
Method: 1. Write the first probability as a fraction and cross-multiply to get equation one. 2. Update the counts after the counters are added, and write equation two in the same way. 3. Simplify both equations and solve them together. 4. Substitute both counts back into the fractions to check them.
After cross-multiplying and expanding, collect everything on one side to get a quadratic. Divide by a common factor first, as that often leaves a quadratic you can factorise by inspection. Factorising is quicker and safer than the formula when the numbers are friendly. Then use the story to choose between the two roots.
A root can be rejected for several reasons. A negative number of counters is impossible. A fraction of a counter is impossible. A condition in the question, such as "there are more than 12 counters", may remove a root that is otherwise valid. Say clearly why you reject it, then state the answer in the context of the question, with units if any. Finally substitute your answer into the original probability to check.
Method: 1. Move all terms to one side and divide by any common factor. 2. Factorise, or use the quadratic formula if it does not factorise. 3. Reject any root that is negative, not a whole number, or breaks a condition. 4. Check the remaining root in the original probability.
Tip: In a "show that" question, the printed quadratic is your target. If your line does not reach it, find the slip in the last line before it.
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