GCSE Maths Probability (Higher): check-up and exam-style questions

Key words. An outcome is one possible result of an experiment. An event is the outcome (or group of outcomes) we are interested in. Equally likely outcomes have the same chance. Probability = number of favourable outcomes ÷ total number of equally likely outcomes, and it is always between 0 and 1.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

1. Find the total number of counters. Every counter is equally likely to be picked, so add up all the counters. This is the bottom of the fraction.
   5 + 3 + 2 = 10
2. Count the counters that are blue. These are the favourable outcomes. This number goes on the top.
   3 blue counters = 3
3. Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
   P(the counter is blue) = 310

Check. The other 7 outcomes do not fit the event, so P(the event does not happen) = 710. Then 310 + 710 = 1010 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Key words. The complement of an event is “the event does not happen”, written not A or A′. Outcomes are exhaustive if together they cover every possibility, so their probabilities add up to 1.

Method. The probabilities of an event happening and not happening add up to 1 (or 100%), so subtract the given probability from 1.

1. Spot the event and its opposite. Either “late” happens or “not late” happens, never both, so these two events are opposites that cover every possibility.
   P(late) = 0.35
2. Write the rule. An event either happens or it does not, so the two probabilities add up to 1. Therefore P(not A) = 1 − P(A).
   P(not late) = 1 − P(late)
3. Subtract from 1. Line up the decimal points and take the probability away from 1.
   1 − 0.35 = 0.65

Check. Add the two probabilities: 0.35 + 0.65 = 1. ✓

Watch out. The answer must be between 0 and 1. If you get a negative number or a number bigger than 1 you have subtracted the wrong way round.

Key words. The complement of an event is “the event does not happen”, written not A or A′. Outcomes are exhaustive if together they cover every possibility, so their probabilities add up to 1.

Method. The probabilities of all the possible outcomes add up to 1. Add the ones we know, then take the total away from 1.

1. Write down what we know. The outcomes listed cover every possibility, so their probabilities add up to 1. One of them is missing.
   P(red) = 12
   P(blue) = 16
   P(green) = ?
2. Add the known probabilities. Use the common denominator 6 so that we can add them.
   36 + 16 = 46 = 23
3. Subtract the total from 1. All the probabilities add up to 1, so the missing one is 1 minus the total of the others.
   66 − 46 = 26 = 13

Check. Add all the probabilities, including the one we found: 12 + 16 + 13 = 1. ✓

Watch out. Do not forget that the missing probability makes the total exactly 1 (not more, not less).

Key words. Frequency is how many times something happened. Relative frequency = frequency ÷ total number of trials. It is an estimate of the probability. An expected number is probability × number of trials.

Method. Relative frequency is the number of times the event happened divided by the number of trials.

1. Write the rule. Relative frequency = frequency ÷ total number of trials. It estimates the probability.
   relative frequency = frequencytrials
2. Put the numbers in. The event happened 18 times in 50 trials.
   1850
3. Simplify the fraction. Divide the top and the bottom by their highest common factor, 2. The value does not change but the numbers are smaller.
   18 ÷ 2 = 9 and 50 ÷ 2 = 25
   1850 = 925

Check. Multiply back: 925 × 50 = 18, which is the frequency 18. ✓

Watch out. The bottom of the fraction is the total number of trials, not the number of times the event did not happen.

Key words. Frequency is how many times something happened. Relative frequency = frequency ÷ total number of trials. It is an estimate of the probability. An expected number is probability × number of trials.

Method. The expected number of times something happens is the probability multiplied by the number of trials.

1. Write down the rule. We multiply how likely the event is on one trial by the number of trials.
   expected number = probability × number of trials
2. Write down the probability and the number of trials. We need both numbers.
   P(red) = 15, number of trials = 150
3. Find the expected number. Expected number = probability × number of trials. Take 15 of 150: divide by 5, then multiply by 1.
   150 ÷ 5 = 30
   30 × 1 = 30

Check. Work backwards: 30 ÷ 150 = 0.2, which is the probability we started with. ✓

Watch out. An expected number is an average over many trials. It does not promise exactly that many will happen.

Key words. Two events are independent if one happening does not change the probability of the other. Mutually exclusive events cannot happen at the same time. For independent events, P(A and B) = P(A) × P(B).

Method. For independent events, “A and B” means we multiply their probabilities: P(A and B) = P(A) × P(B).

1. Check that the events are independent. The outcome of one does not change the chances for the other. So we can multiply their probabilities.
   Independent events
2. Write down the probability of each event. We need each probability on its own.
   P(first coin heads) = 12
   P(second coin heads) = 12
3. Multiply the probabilities. For independent events, P(A and B) = P(A) × P(B). Multiply the tops together and the bottoms together.
   12 × 12 = 14

Check. Both things have to happen, so the answer must be smaller than each single probability: 14 is smaller than 12. ✓

Watch out. Multiply for “and”. Do not add the probabilities.

Key words. An outcome is one possible result of an experiment. An event is the outcome (or group of outcomes) we are interested in. Equally likely outcomes have the same chance. Probability = number of favourable outcomes ÷ total number of equally likely outcomes, and it is always between 0 and 1.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

1. Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.

   +	1	2	3	4	5	6
   1	2	3	4	5	6	7
   2	3	4	5	6	7	8
   3	4	5	6	7	8	9
   4	5	6	7	8	9	10
   5	6	7	8	9	10	11
   6	7	8	9	10	11	12

   Total number of outcomes = 6 × 6 = 36
2. Highlight the cells where the total is 9. These are the favourable outcomes. Their number goes on the top of the fraction.

   +	1	2	3	4	5	6
   1	2	3	4	5	6	7
   2	3	4	5	6	7	8
   3	4	5	6	7	8	9
   4	5	6	7	8	9	10
   5	6	7	8	9	10	11
   6	7	8	9	10	11	12

   Number of highlighted cells = 4
3. Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
   P(the total is 9) = 436
4. Simplify the fraction. Divide the top and the bottom by their highest common factor, 4. The value does not change but the numbers are smaller.
   4 ÷ 4 = 1 and 36 ÷ 4 = 9
   436 = 19

Check. The other 32 outcomes do not fit the event, so P(the event does not happen) = 89. Then 436 + 3236 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Key words. A tree diagram shows every outcome of two or more events in order. The probabilities on the branches from one point always add to 1. Multiply along a route (“and”) and add the routes you want (“or”).

Method. Find the routes through the tree that give the event, multiply the probabilities along each route, then add the routes together.

1. Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
   red then red
2. Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
   red then red: 35 × 24 = 620 = 310

Check. The routes not used add up to 710, and 310 + 710 = 1. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Key words. Conditional probability is the probability of an event given that another event has happened. The word “given” shrinks the sample space to just the group we are told about. P(A | B) = n(A ∩ B) ÷ n(B).

Method. For a conditional probability, shrink the sample space to the group we are told about (the “given” group), then work out probability = favourable ÷ size of that group.

1. Narrow down to the group we are told about: the student is a girl. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.
   Girls row total = 9 + 11 = 20
2. Count how many in that group also have the event: voted for the zoo. This is the number we want. It goes on the top.
   9 girls voted for the zoo: 9
3. Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
   P(voted for the zoo | the student is a girl) = 920

Check. The other 11 outcomes do not fit the event, so P(the event does not happen) = 1120. Then 920 + 1120 = 2020 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. The bottom of a conditional probability is the size of the group after “given”, not the total of everyone.

Method. Write P(red) with x, put it equal to 2/3 and solve.

1. Write P(red). Red over all counters.
   xx + 4 = 23
2. Cross-multiply. Multiply both sides by 3(x + 4).
   3x = 2(x + 4) = 2x + 8
3. Solve. Subtract 2x.
   x = 8

Check. 8 red and 4 blue: P(red) = 8/12 = 2/3. ✓

Watch out. Blue counters are part of the total.

Key words. An outcome is one possible result of an experiment. An event is the outcome (or group of outcomes) we are interested in. Equally likely outcomes have the same chance. Probability = number of favourable outcomes ÷ total number of equally likely outcomes, and it is always between 0 and 1.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

1. Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.

   ×	1	2	3	4	5	6
   1	1	2	3	4	5	6
   2	2	4	6	8	10	12
   3	3	6	9	12	15	18
   4	4	8	12	16	20	24
   5	5	10	15	20	25	30
   6	6	12	18	24	30	36

   Total number of outcomes = 6 × 6 = 36
2. Highlight the cells where the product is 12. These are the favourable outcomes. Their number goes on the top of the fraction.

   ×	1	2	3	4	5	6
   1	1	2	3	4	5	6
   2	2	4	6	8	10	12
   3	3	6	9	12	15	18
   4	4	8	12	16	20	24
   5	5	10	15	20	25	30
   6	6	12	18	24	30	36

   Number of highlighted cells = 4
3. Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
   P(the product is 12) = 436
4. Simplify the fraction. Divide the top and the bottom by their highest common factor, 4. The value does not change but the numbers are smaller.
   4 ÷ 4 = 1 and 36 ÷ 4 = 9
   436 = 19

Check. The other 32 outcomes do not fit the event, so P(the event does not happen) = 89. Then 436 + 3236 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Key words. Two events are independent if one happening does not change the probability of the other. Mutually exclusive events cannot happen at the same time. For independent events, P(A and B) = P(A) × P(B).

Method. For independent events, “A and B” means we multiply their probabilities: P(A and B) = P(A) × P(B).

1. Check that the events are independent. The outcome of one does not change the chances for the other. So we can multiply their probabilities.
   Independent events
2. Write down the probability of each event. We need each probability on its own.
   P(heads) = 12
   P(an even number) = 25
3. Multiply the probabilities. For independent events, P(A and B) = P(A) × P(B). Multiply the tops together and the bottoms together.
   12 × 25 = 210 = 15

Check. Both things have to happen, so the answer must be smaller than each single probability: 15 is smaller than 25. ✓

Watch out. Multiply for “and”. Do not add the probabilities.

Key words. A tree diagram shows every outcome of two or more events in order. The probabilities on the branches from one point always add to 1. Multiply along a route (“and”) and add the routes you want (“or”).

Method. Find the routes through the tree that give the event, multiply the probabilities along each route, then add the routes together.

1. Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
   red then blue
   blue then red
2. Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
   red then blue: 410 × 69 = 2490 = 415
   blue then red: 610 × 49 = 2490 = 415
3. Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
   415 + 415 = 815

Check. The routes not used add up to 715, and 815 + 715 = 1. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Method. Build the frequency tree from the fractions, then add the vegetarian branches.

1. Adults and children. Multiply the total by each fraction.
   Adults: 23 of 120 = 80
   Children: 120 − 80 = 40
2. Vegetarians in each group. Take the fraction of each group.
   Adults: 14 of 80 = 20
   Children: 12 of 40 = 20
3. Add the vegetarians. Add the two vegetarian branches.
   20 + 20 = 40
4. Show the tree. Write the numbers in the boxes.

Check. Not vegetarian: 60 + 20 = 80, and 40 + 80 = 120. ✓

Watch out. Marks: 2 for the first branches, 1 for the second branches, 1 for the answer.

Key words. Two events are independent if one happening does not change the probability of the other. Mutually exclusive events cannot happen at the same time. For independent events, P(A and B) = P(A) × P(B).

Method. P(at least one) = 1 − P(none). Find the probability of none by multiplying, then subtract from 1.

1. Use the opposite event. “At least one late day” is the opposite of “on time every time”. It is easier to find the opposite and subtract from 1.
   P(at least one) = 1 − P(none)
2. Find the probability of the event not happening on one trial. Use P(not A) = 1 − P(A).
   P(not late day) = 1 − 0.1 = 0.9
3. Multiply for all 3 trials. The trials are independent, so multiply the 3 probabilities.
   P(none) = 0.9 × 0.9 × 0.9 = 0.729
4. Subtract from 1. P(at least one) = 1 − P(none).
   1 − 0.729 = 0.271

Check. P(none) + P(at least one) = 0.729 + 0.271 = 1. ✓

Watch out. Do not add the probabilities of each trial. For “at least one” always think “1 minus none”.

Method. The rule favourable ÷ total needs equally likely outcomes. List the outcomes properly.

1. List all outcomes. The coins are separate: first then second.
   HH, HT, TH, TT: 4 equally likely outcomes
2. Correct the probability. Only HH has two heads.
   P(two heads) = 14
3. Explain Sam’s mistake. One head can happen in two ways (HT and TH), so “one head” is not as unlikely as the others.
   The three outcomes are not equally likely.

Check. 1/4 + 1/2 + 1/4 = 1. ✓

Watch out. Marks: 1 for listing four outcomes, 1 for the explanation.

Method. Find the expected number of winners, then money in minus money out.

1. Expected winners. Probability × number of plays.
   18 × 400 = 50
2. Money in. Each play costs £2.
   400 × £2 = £800
3. Prizes out. Each winner gets £12.
   50 × £12 = £600
4. Profit. Money in minus money out.
   £800 − £600 = £200

Check. Each play: £2 − 1/8 × £12 = £0.50 profit, and 400 × £0.50 = £200. ✓

Watch out. Marks: 1 expected winners, 1 money in, 1 prizes, 1 profit.

Key words. Conditional probability is the probability of an event given that another event has happened. The word “given” shrinks the sample space to just the group we are told about. P(A | B) = n(A ∩ B) ÷ n(B).

Method. P(A | B) = P(A and B) ÷ P(B). Use the tree to find P(B) and P(A and B), then divide.

1. Write the rule. Conditional probability: the probability of the event given the condition is the probability of both, divided by the probability of the condition.
   P(first red | second blue) = P(first red and second blue)P(second blue)
2. Find P(second blue). Find every route on the tree where the condition happens, multiply along each, and add.
   red then blue: 610 × 49 = 2490 = 415
   blue then blue: 410 × 39 = 1290 = 215
   415 + 215 = 615 = 25
   P(second blue) = 25
3. Find P(first red and second blue). Of those routes, pick the ones where the event also happens.
   red then blue: 610 × 49 = 2490 = 415
   P(first red and second blue) = 415
4. Divide. Write both probabilities over a common denominator, so we can just divide the numerators.
   415 ÷ 615 = 4 ÷ 6 = 23

Check. Think of 6 equally likely cases where the condition holds, of which 4 also have the event: 23. ✓

Watch out. The bottom is the probability of the given condition, not 1 and not the total.

Method. Write P(both red) in terms of x, equate to 2/15, rearrange and factorise.

1. Write the probabilities. After a red counter is taken: 3 red and 3 + x counters left.
   P(1st red) = 44 + x, P(2nd red) = 33 + x
2. Multiply and equate. Both red means first and second.
   12(4 + x)(3 + x) = 215
3. Cross-multiply. Multiply both sides by 15 and by the bottom, then divide by 2.
   180 = 2(4 + x)(3 + x), so 90 = (4 + x)(3 + x)
4. Expand. (4 + x)(3 + x) = 12 + 7x + x².
   x² + 7x + 12 = 90
5. Rearrange. Subtract 90 from both sides.
   x² + 7x − 78 = 0
6. Factorise and solve. Numbers that multiply to −78 and add to 7 are 13 and −6. A number of counters is positive.
   (x + 13)(x − 6) = 0, so x = 6

Check. 6 blue and 4 red: 4/10 × 3/9 = 12/90 = 2/15. ✓

Watch out. Marks: 1 for each probability, 1 for the equation, 1 for the quadratic, 1 for the solution.