GCSE Maths: Conditional probability and probability with algebra (Higher)
Part of the GCSE Maths probability chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Conditional probability and probability with algebra
Sometimes we are told something has already happened, and we are asked how likely something else is given that. This is conditional probability. It is written P(A | B), read “the probability of A given B”. For example, “P(girl | walks)” means: choose a student who walks, and ask how likely it is that this student is a girl.
The word given shrinks the sample space. We are no longer choosing from everyone. We are choosing only from the group we are told about. So the bottom of the probability fraction is the size of that group, not the grand total.
| Walk | Bus | Total | |
|---|---|---|---|
| Boys | 12 | 6 | 18 |
| Girls | 9 | 8 | 17 |
| Total | 21 | 14 | 35 |
From a two-way table: restrict to the row or column named after “given”. From a Venn diagram: P(A | B) = n(A ∩ B) ÷ n(B). From a tree diagram: use the rule P(A | B) = P(A and B) ÷ P(B), where P(A and B) is a route probability and P(B) is found by adding the routes where B happens. On a tree with dependent events the branch for the second stage is already a conditional probability: for example, the branch probability 3/9 on the second stage after a red is P(second red | first red).
Order matters. P(A | B) is usually not the same as P(B | A). The probability that a person who owns a pet is a child is different from the probability that a child owns a pet. Independence. Events A and B are independent exactly when knowing B does not change the probability of A, that is P(A | B) = P(A), which is the same as P(A and B) = P(A) × P(B).
Algebra and probability. Probabilities can contain unknowns. If a bag has x red counters and 5 blue counters, then P(red) = x/(x + 5). If we are told P(red) = 3/5, we form an equation and solve it. We use two facts again and again: all the probabilities of the outcomes add up to 1, and probabilities of picks without replacement multiply with the numbers going down by one.
Key ideas
- P(A | B) means the probability of A given that B has happened. “Given” shrinks the sample space to the group B.
- Table or Venn: P(A | B) = n(A ∩ B) ÷ n(B). Tree: P(A | B) = P(A and B) ÷ P(B).
- P(A | B) is not usually the same as P(B | A).
- A and B are independent when P(A | B) = P(A), that is P(A and B) = P(A) × P(B).
- With algebra: write each probability as a fraction containing the unknown, make an equation, and solve it. All probabilities add up to 1.
| Walk | Bus | Total | |
|---|---|---|---|
| Boys | 12 | 6 | 18 |
| Girls | 9 | 8 | 17 |
| Total | 21 | 14 | 35 |
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.
- Narrow down to the group we are told about: the student walks. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.Walk column total = 12 + 9 = 21 students
- Count how many in that group also have the event: the student is a girl. This is the number we want. It goes on the top.9 of them are girls, so 9
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(the student is a girl | the student walks) = 921
- Simplify the fraction. Divide the top and the bottom by their highest common factor, 3. The value does not change but the numbers are smaller.9 ÷ 3 = 3 and 21 ÷ 3 = 7
921 = 37
Check. The other 12 outcomes do not fit the event, so P(the event does not happen) = 47. Then 921 + 1221 = 2121 = 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.
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.
- Narrow down to the group we are told about: B. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.n(B) = 5 + 7 = 12 students study music
- Count how many in that group also have the event: A. This is the number we want. It goes on the top.n(A ∩ B) = 5 students are in both
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(A | B) = 512
Check. The other 7 outcomes do not fit the event, so P(the event does not happen) = 712. Then 512 + 712 = 1212 = 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.
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.
- 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 red) = P(first red and second red)P(second red)
- Find P(second red). Find every route on the tree where the condition happens, multiply along each, and add.red then red: 410 × 39 = 1290 = 215
blue then red: 610 × 49 = 2490 = 415
215 + 415 = 615 = 25
P(second red) = 25 - Find P(first red and second red). Of those routes, pick the ones where the event also happens.red then red: 410 × 39 = 1290 = 215
P(first red and second red) = 215 - Divide. Write both probabilities over a common denominator, so we can just divide the numerators.215 ÷ 615 = 2 ÷ 6 = 13
Check. Think of 6 equally likely cases where the condition holds, of which 2 also have the event: 13. ✓
Watch out. The bottom is the probability of the given condition, not 1 and not the total.
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. Use expected numbers (a frequency tree). The “given positive” group is everyone with a positive test, so find its size.
- Find the number with the condition. 2% of 1000.0.02 × 1000 = 20 people have the condition, 980 do not.
- Find the positive tests in each group. 90% of 20 and 5% of 980.0.9 × 20 = 18
0.05 × 980 = 49 - Find the total number of positive tests. This is the size of the group we are told about (“given positive”).18 + 49 = 67
- Find how many of these have the condition. Positive tests among those with the condition.18 people
- Write the probability. Number with the condition and positive, over all positives.1867
- Show the frequency tree. This summarises all the numbers.
Check. 18 ÷ 67 = 0.269: only about 27%, even though the test looks accurate, because the condition is rare. ✓
Watch out. The bottom is all the positive tests, not all 1000 people and not just the 20 who have the condition.
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. Use P(A | B) = P(A and B) ÷ P(B), then compare with P(A).
- Write the rule. Conditional probability formula.P(A | B) = P(A and B) ÷ P(B)
- Substitute. Use the given values.0.2 ÷ 0.5 = 2 ÷ 5 = 0.4
- Compare with P(A). If P(A | B) = P(A), knowing B does not change the chance of A.P(A | B) = 0.4 = P(A)
- Conclude. A and B are independent.
Check. P(A) × P(B) = 0.4 × 0.5 = 0.2 = P(A and B). ✓
Watch out. Independence is tested by P(A | B) = P(A).
Method. Write P(blue) as a fraction containing x, put it equal to 5/8, and solve.
- Count all the counters. The total is red plus blue.Total = 3 + x
- Write P(blue). Blue counters over all counters.P(blue) = x3 + x
- Form the equation. We are told P(blue) = 5/8.x3 + x = 58
- Cross-multiply. Multiply both sides by 8 and by (3 + x) to clear the fractions.8x = 5(3 + x)
- Expand and solve. Multiply out the bracket, subtract 5x from both sides, then divide by 3.8x = 15 + 5x
3x = 15
x = 5
Check. With 5 blue and 3 red there are 8 counters and P(blue) = 5/8. ✓
Watch out. Remember that the number of blue counters is also part of the total on the bottom.
Method. Write P(both red) as a product of two fractions in terms of x, equate to 2/9, rearrange into a quadratic and solve by factorising.
- Write the two probabilities. After the first red counter is taken, there is one fewer red and one fewer counter in total.P(1st red) = 55 + x, P(2nd red) = 44 + x
- Multiply and set equal to 2/9. Both red means first and second.55 + x × 44 + x = 20(5 + x)(4 + x) = 29
- Cross-multiply. Multiply both sides by 9 and by the bottom, then divide by 2.20 × 9 = 2(5 + x)(4 + x)
90 = (5 + x)(4 + x) - Expand and rearrange. (5 + x)(4 + x) = 20 + 9x + x2. Subtract 90 from both sides.20 + 9x + x2 = 90
x2 + 9x − 70 = 0 - Factorise and solve. Two numbers that multiply to −70 and add to 9 are 14 and −5.(x + 14)(x − 5) = 0
x = −14 or x = 5 - Choose the sensible answer. The number of counters cannot be negative.x = 5
Check. 5 red and 5 blue: P(both red) = 5/10 × 4/9 = 20/90 = 2/9. ✓
Watch out. Reject the negative solution in context.
Watch out
P(A | B) and P(B | A) are different. And the bottom of a conditional probability is the size of the given group: 9/21 for “a girl, given walks”, not 9/35. With algebra, do not forget that a counter taken without replacement also reduces the total: both the top and the bottom change.Exercise 23M Conditional probability
Non-calculator| Walk | Bus | Total | |
|---|---|---|---|
| Boys | 12 | 6 | 18 |
| Girls | 9 | 8 | 17 |
| Total | 21 | 14 | 35 |
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.
- Narrow down to the group we are told about: takes the bus. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.Bus column total = 6 + 8 = 14
- Count how many in that group also have the event: a boy. This is the number we want. It goes on the top.6 boys take the bus: 6
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(a boy | takes the bus) = 614
- 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.6 ÷ 2 = 3 and 14 ÷ 2 = 7
614 = 37
Check. The other 8 outcomes do not fit the event, so P(the event does not happen) = 47. Then 614 + 814 = 1414 = 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.
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.
- Narrow down to the group we are told about: 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 + 8 = 17
- Count how many in that group also have the event: walks. This is the number we want. It goes on the top.9 girls walk: 9
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(walks | is a girl) = 917
Check. The other 8 outcomes do not fit the event, so P(the event does not happen) = 817. Then 917 + 817 = 1717 = 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.
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.
- Narrow down to the group we are told about: is a boy. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.Boys row total = 12 + 6 = 18
- Count how many in that group also have the event: takes the bus. This is the number we want. It goes on the top.6 boys take the bus: 6
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(takes the bus | is a boy) = 618
- Simplify the fraction. Divide the top and the bottom by their highest common factor, 6. The value does not change but the numbers are smaller.6 ÷ 6 = 1 and 18 ÷ 6 = 3
618 = 13
Check. The other 12 outcomes do not fit the event, so P(the event does not happen) = 23. Then 618 + 1218 = 1818 = 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.
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.
- Narrow down to the group we are told about: B. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.n(B) = 4 + 6 = 10
- Count how many in that group also have the event: A. This is the number we want. It goes on the top.n(A ∩ B) = 4
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(A | B) = 410
- 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.4 ÷ 2 = 2 and 10 ÷ 2 = 5
410 = 25
Check. The other 6 outcomes do not fit the event, so P(the event does not happen) = 35. Then 410 + 610 = 1010 = 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.
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.
- Narrow down to the group we are told about: A. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.n(A) = 9 + 4 = 13
- Count how many in that group also have the event: B. This is the number we want. It goes on the top.n(A ∩ B) = 4
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(B | A) = 413
Check. The other 9 outcomes do not fit the event, so P(the event does not happen) = 913. Then 413 + 913 = 1313 = 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.
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.
- Narrow down to the group we are told about: B′. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.n(B′) = 9 + 3 = 12
- Count how many in that group also have the event: A. This is the number we want. It goes on the top.n(A ∩ B′) = 9
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(A | B′) = 912
- Simplify the fraction. Divide the top and the bottom by their highest common factor, 3. The value does not change but the numbers are smaller.9 ÷ 3 = 3 and 12 ÷ 3 = 4
912 = 34
Check. The other 3 outcomes do not fit the event, so P(the event does not happen) = 14. Then 912 + 312 = 1212 = 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.
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.
- Narrow down to the group we are told about: the score is even. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.Even scores: 2, 4, 6, so 3
- Count how many in that group also have the event: the score is greater than 3. This is the number we want. It goes on the top.Of these, 4 and 6 are greater than 3: 2
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(the score is greater than 3 | the score is even) = 23
Check. The other 1 outcome do not fit the event, so P(the event does not happen) = 13. Then 23 + 13 = 33 = 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.
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.
- Narrow down to the group we are told about: a heart. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.There are 13 hearts: 13
- Count how many in that group also have the event: a king. This is the number we want. It goes on the top.Only the king of hearts is both: 1
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(a king | a heart) = 113
Check. The other 12 outcomes do not fit the event, so P(the event does not happen) = 1213. Then 113 + 1213 = 1313 = 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. Once the first counter is red, there are 3 red counters among 9 left.
- Remove the first counter. It was red, so 4 − 1 = 3 red remain and 9 counters in total remain.3 red out of 9
- Write the probability. Red left over all left.39 = 13
Check. This is the second-stage branch after red on the tree. ✓
Watch out. Use the numbers left in the bag.
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.
- 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 red) = P(first red and second red)P(second red)
- Find P(second red). Find every route on the tree where the condition happens, multiply along each, and add.red then red: 410 × 39 = 1290 = 215
blue then red: 610 × 49 = 2490 = 415
215 + 415 = 615 = 25
P(second red) = 25 - Find P(first red and second red). Of those routes, pick the ones where the event also happens.red then red: 410 × 39 = 1290 = 215
P(first red and second red) = 215 - Divide. Write both probabilities over a common denominator, so we can just divide the numerators.215 ÷ 615 = 2 ÷ 6 = 13
Check. Think of 6 equally likely cases where the condition holds, of which 2 also have the event: 13. ✓
Watch out. The bottom is the probability of the given condition, not 1 and not the total.
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.
- 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(different colours | second blue) = P(different colours and second blue)P(second blue)
- Find P(second blue). Find every route on the tree where the condition happens, multiply along each, and add.red then blue: 410 × 69 = 2490 = 415
blue then blue: 610 × 59 = 3090 = 13
415 + 13 = 415 + 515 = 915 = 35
P(second blue) = 35 - Find P(different colours and second blue). Of those routes, pick the ones where the event also happens.red then blue: 410 × 69 = 2490 = 415
P(different colours and second blue) = 415 - Divide. Write both probabilities over a common denominator, so we can just divide the numerators.415 ÷ 915 = 4 ÷ 9 = 49
Check. Think of 9 equally likely cases where the condition holds, of which 4 also have the event: 49. ✓
Watch out. The bottom is the probability of the given condition, not 1 and not the total.
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.
- Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.rain then rain
- Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.rain then rain: 0.3 × 0.6 = 0.18
Check. The routes not used add up to 0.82, and 0.18 + 0.82 = 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. 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.
- Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.rain then rain
dry then rain - Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.rain then rain: 0.3 × 0.6 = 0.18
dry then rain: 0.7 × 0.2 = 0.14 - Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.0.18 + 0.14 = 0.32
Check. The routes not used add up to 0.68, and 0.32 + 0.68 = 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. P(A | B) = P(A and B) ÷ P(B). Use the tree to find P(B) and P(A and B), then divide.
- 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(rain Monday | rain Tuesday) = P(rain Monday and rain Tuesday)P(rain Tuesday)
- Find P(rain Tuesday). Find every route on the tree where the condition happens, multiply along each, and add.rain then rain: 0.3 × 0.6 = 0.18
dry then rain: 0.7 × 0.2 = 0.14
0.18 + 0.14 = 0.32
P(rain Tuesday) = 0.32 - Find P(rain Monday and rain Tuesday). Of those routes, pick the ones where the event also happens.rain then rain: 0.3 × 0.6 = 0.18
P(rain Monday and rain Tuesday) = 0.18 - Divide. Write both probabilities over a common denominator, so we can just divide the numerators.950 ÷ 1650 = 9 ÷ 16 = 916
Check. Think of 16 equally likely cases where the condition holds, of which 9 also have the event: 916. ✓
Watch out. The bottom is the probability of the given condition, not 1 and not the total.
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. Use expected numbers on a frequency tree. The group we are told about is everyone with a positive test.
- Find the people with and without the disease. 1% of 2000 have it.0.01 × 2000 = 20 have the disease, 1980 do not
- Find the positive tests in each group. 95% of 20 and 4% of 1980.0.95 × 20 = 19
0.04 × 1980 = 79.2 - Add the positive tests. This is the size of the group we are told about.19 + 79.2 = 98.2
- Write the probability. Positive and diseased over all positive.19 ÷ 98.2 = 0.19 to 2 d.p.
Check. Most positive tests are false alarms: 79.2 out of 98.2. This is because the disease is rare. ✓
Watch out. Use all the positive tests as the bottom, not 2000 and not 20.
Method. Work out both conditional probabilities and compare.
- Find P(A | B). Overlap over n(B).410 = 25
- Find P(B | A). Overlap over n(A).413
- Compare. 2/5 = 0.4 and 4/13 = 0.31 (to 2 d.p.). They are different.P(A | B) ≠ P(B | A)
Check. They share the same top (4) but have different bottoms (10 and 13), so they are equal only when n(A) = n(B). ✓
Watch out. The bottom number is the “given” group, which is different each time.
Method. Independent events have P(A | B) = P(A). Find P(A | B) and compare.
- Write the rule. P(A | B) = P(A and B) ÷ P(B).0.1 ÷ 0.4
- Work it out. Multiply top and bottom by 10.1 ÷ 4 = 0.25
- Compare with P(A). P(A) = 0.5.0.25 ≠ 0.5
- Conclude. Knowing B changes the probability of A, so they are not independent.A and B are not independent.
Check. P(A) × P(B) = 0.5 × 0.4 = 0.2, which is not P(A and B) = 0.1. ✓
Watch out. Check independence with either rule, and state clearly which one you use.
| Instrument | None | Total | |
|---|---|---|---|
| Year 10 | 16 | 14 | 30 |
| Year 11 | 24 | 26 | 50 |
| Total | 40 | 40 | 80 |
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.
- Narrow down to the group we are told about: plays an instrument. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.Instrument column total = 16 + 24 = 40
- Count how many in that group also have the event: Year 10. This is the number we want. It goes on the top.16 are in Year 10: 16
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(Year 10 | plays an instrument) = 1640
- Simplify the fraction. Divide the top and the bottom by their highest common factor, 8. The value does not change but the numbers are smaller.16 ÷ 8 = 2 and 40 ÷ 8 = 5
1640 = 25
Check. The other 24 outcomes do not fit the event, so P(the event does not happen) = 35. Then 1640 + 2440 = 4040 = 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. Use the grand total, because no condition is given.
- Find the Year 10 total. Row total.30
- Divide by the grand total. 3080 = 38
Check. 3/8 = 0.375. ✓
Watch out. Without “given” the bottom is everyone.
Method. Compare the two probabilities.
- Change to decimals. P(Year 10 | instrument) = 16/40 and P(Year 10) = 30/80.0.4 and 0.375
- Compare. 0.4 is bigger than 0.375.P(Year 10 | instrument) > P(Year 10)
- Interpret. Knowing that a student plays an instrument changes the probability of Year 10, so the events are not independent.Not independent
Check. If they were independent, both probabilities would be 0.375. ✓
Watch out. Independence needs the two probabilities to be exactly equal.
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.
- 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)
- Find P(second blue). Find every route on the tree where the condition happens, multiply along each, and add.red then blue: 38 × 57 = 1556
blue then blue: 58 × 47 = 2056 = 514
1556 + 514 = 1556 + 2056 = 3556 = 58
P(second blue) = 58 - Find P(first red and second blue). Of those routes, pick the ones where the event also happens.red then blue: 38 × 57 = 1556
P(first red and second blue) = 1556 - Divide. Write both probabilities over a common denominator, so we can just divide the numerators.1556 ÷ 3556 = 15 ÷ 35 = 37
Check. Think of 35 equally likely cases where the condition holds, of which 15 also have the event: 37. ✓
Watch out. The bottom is the probability of the given condition, not 1 and not the total.
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.
- Narrow down to the group we are told about: B. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.n(B) = 5 + 3 + 4 + 1 = 13
- Count how many in that group also have the event: A. This is the number we want. It goes on the top.n(A ∩ B) = 3 + 1 = 4
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(A | B) = 413
Check. The other 9 outcomes do not fit the event, so P(the event does not happen) = 913. Then 413 + 913 = 1313 = 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.
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.
- Narrow down to the group we are told about: A′. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.n(A′) = 27 (everything outside circle A)
- Count how many in that group also have the event: C. This is the number we want. It goes on the top.n(C ∩ A′) = 10 (the parts of C outside A)
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(C | A′) = 1027
Check. The other 17 outcomes do not fit the event, so P(the event does not happen) = 1727. Then 1027 + 1727 = 2727 = 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.
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.
- Narrow down to the group we are told about: C. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.n(C) = 13
- Count how many in that group also have the event: A ∩ B. This is the number we want. It goes on the top.n(A ∩ B ∩ C) = 1
- Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.P(A ∩ B | C) = 113
Check. The other 12 outcomes do not fit the event, so P(the event does not happen) = 1213. Then 113 + 1213 = 1313 = 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.
Exercise 23N Probability and algebra
Non-calculatorMethod. Write P(blue) with x, make it equal to 5/8 and solve.
- Write P(blue). Blue over all counters (3 + x).x3 + x
- Form the equation. It equals 5/8.x3 + x = 58
- Cross-multiply. Multiply both sides by 8(3 + x).8x = 5(3 + x) = 15 + 5x
- Solve. Subtract 5x then divide by 3.3x = 15, so x = 5
Check. 5 blue and 3 red: P(blue) = 5/8. ✓
Watch out. The unknown is part of the total.
Method. The total is a sum with x. Write P(green) and solve.
- Find the total. Add red, blue and green.x + 2x + 6 = 3x + 6
- Write P(green). Green over the total.63x + 6 = 14
- Cross-multiply. Multiply both sides by 4(3x + 6).24 = 3x + 6
- Solve. Subtract 6, then divide by 3.18 = 3x, so x = 6
- Find the number of counters. Put x = 6 into 3x + 6.3 × 6 + 6 = 24 counters
Check. 6 red, 12 blue, 6 green = 24, and 6/24 = 1/4. ✓
Watch out. Do not forget that the 2x blue counters count towards the total.
Method. The probabilities add up to 1. Form and solve an equation.
- Write the equation. Add all four probabilities and set the total to 1.0.3 + x + 2x + (x + 0.1) = 1
- Collect like terms. Add the x terms and the numbers.4x + 0.4 = 1
- Subtract 0.4 from both sides. Get the x term on its own.4x = 0.6
- Divide by 4. Find one x.x = 0.6 ÷ 4 = 0.15
- Find P(D). D has probability x + 0.1.0.15 + 0.1 = 0.25
Check. 0.3 + 0.15 + 0.3 + 0.25 = 1. ✓
Watch out. Always check that every probability is between 0 and 1.
Method. Write P(both red) in terms of n, equate to 1/11, rearrange into a quadratic and factorise.
- Write the two probabilities. After one red counter is taken there are 3 red and n − 1 counters left.P(1st red) = 4n, P(2nd red) = 3n − 1
- Multiply and set equal to 1/11. Both red means first and second.12n(n − 1) = 111
- Cross-multiply. Multiply both sides by 11 and by n(n − 1).132 = n(n − 1) = n2 − n
- Rearrange to equal zero. Subtract 132 from both sides.n2 − n − 132 = 0
- Factorise. Two numbers that multiply to −132 and add to −1 are −12 and 11.(n − 12)(n + 11) = 0
- Solve and reject the negative. A number of counters cannot be negative.n = 12 (or −11, rejected)
Check. 12 counters, 4 red: 4/12 × 3/11 = 12/132 = 1/11. ✓
Watch out. Reject the negative solution because n is a number of counters.
Method. With replacement, both picks have the same probability. Square it, set it equal to 1/4, and take the square root.
- Write P(red). Red over all counters.P(red) = xx + 6
- Write P(both red). With replacement the picks are independent: multiply.xx + 6 × xx + 6 = x2(x + 6)2 = 14
- Take square roots. A probability is positive, so use the positive root of 1/4, which is 1/2.xx + 6 = 12
- Solve. Cross-multiply: 2x = x + 6.2x = x + 6, so x = 6
Check. 6 red and 6 blue: P(red) = 1/2, and 1/2 × 1/2 = 1/4. ✓
Watch out. Take the square root of both sides before cross-multiplying, to keep the algebra simple.
Method. Write P(both red) in terms of x, equate to 1/5, and solve the quadratic.
- Write the two probabilities. After one red is taken: 2 red and x + 2 counters left.P(1st red) = 33 + x, P(2nd red) = 22 + x
- Multiply and set equal to 1/5. Both red means first and second.6(3 + x)(2 + x) = 15
- Cross-multiply. Multiply both sides by 5 and by the bottom.30 = (3 + x)(2 + x) = 6 + 5x + x2
- Rearrange. Subtract 30 from both sides.x2 + 5x − 24 = 0
- Factorise. Numbers that multiply to −24 and add to 5: 8 and −3.(x + 8)(x − 3) = 0
- Solve. The number of counters must be positive.x = 3
Check. 3 red and 3 blue: 3/6 × 2/5 = 6/30 = 1/5. ✓
Watch out. The second probability has 2 added to x on the bottom, not 3.
Method. All the regions add up to 40. Form an equation, solve it, then use the overlap.
- Add the regions. The four numbers must add up to the total.(x + 3) + x + 2x + 5 = 40
- Collect like terms. Add the x terms and the numbers.4x + 8 = 40
- Solve. Subtract 8, then divide by 4.4x = 32, so x = 8
- Find the overlap. A ∩ B has x students.8 students
- Write the probability. Overlap over the total.840 = 15
Check. Regions: 11 + 8 + 16 + 5 = 40. ✓
Watch out. Use all four regions, including the one outside the circles.
Method. Use P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and solve for x.
- Write the formula with the values. Put in each probability.0.5 = x + 2x − 0.1
- Collect like terms. x + 2x = 3x.0.5 = 3x − 0.1
- Add 0.1 to both sides. This isolates the x term.0.6 = 3x
- Divide by 3. Find one x.x = 0.2
Check. P(A) = 0.2, P(B) = 0.4, so 0.2 + 0.4 − 0.1 = 0.5. ✓
Watch out. Do not forget to subtract the overlap.
Method. Both days: multiply p by itself. Solve p2 = 0.09, then use the tree.
- Form the equation. Both late means late and late, so multiply.p × p = 0.09, so p2 = 0.09
- Take the square root. A probability cannot be negative.p = 0.3
- Find P(not late). 1 minus P(late).1 − 0.3 = 0.7
- Find the two routes for exactly one late day. Late then on time, or on time then late.0.3 × 0.7 = 0.21 and 0.7 × 0.3 = 0.21
- Add the routes. Different routes: add.0.21 + 0.21 = 0.42
Check. 0.09 + 0.42 + 0.49 = 1. ✓
Watch out. Reject the negative square root.
Method. Call the original total T. Red is 3/5 of T, and the red number stays the same after adding blue.
- Write the red counters in terms of T. P(red) = 3/5.red = 35 T
- Write red after adding 4 blue. The total is now T + 4 and the red number is unchanged.red = 12(T + 4)
- Form an equation. Both expressions equal the number of red counters.35 T = 12(T + 4)
- Multiply both sides by 10. To clear the fractions.6T = 5(T + 4) = 5T + 20
- Solve. Subtract 5T.T = 20
- Find each colour. Red = 3/5 of 20, blue is the rest.red = 12, blue = 20 − 12 = 8
Check. After adding 4 blue: 12 red out of 24, which is 1/2. ✓
Watch out. The number of red counters does not change when blue counters are added.
Method. Put the numbers into P(A | B) = P(A and B) ÷ P(B) and solve.
- Write the rule. Conditional probability.0.4 = 0.12 ÷ x
- Multiply both sides by x. To remove x from the bottom.0.4x = 0.12
- Divide by 0.4. Multiply top and bottom by 10 to get whole numbers.x = 0.12 ÷ 0.4 = 1.2 ÷ 4 = 0.3
Check. 0.12 ÷ 0.3 = 0.4. ✓
Watch out. x is on the bottom, so multiply both sides by x first.
Method. Find P(win second game) from the two routes on a tree, in terms of p, and equate to 0.55.
- Find the two routes that give a win in the second game. Win then win, or lose then win.P(lose first) = 1 − p
- Multiply along each route. And means multiply.win, win: 0.7p
lose, win: 0.4(1 − p) - Add the routes and form an equation. Different routes: add.0.7p + 0.4(1 − p) = 0.55
- Expand and collect. 0.4(1 − p) = 0.4 − 0.4p.0.7p + 0.4 − 0.4p = 0.55
0.3p + 0.4 = 0.55 - Solve. Subtract 0.4, then divide by 0.3.0.3p = 0.15, so p = 0.5
Check. 0.7 × 0.5 + 0.4 × 0.5 = 0.35 + 0.2 = 0.55. ✓
Watch out. Do not forget the route where she loses the first game and wins the second.
Method. There are two routes: red then blue, and blue then red. Find each in terms of x and y and add.
- Find the total counters. x red and y blue.Total = x + y; after one pick, x + y − 1
- Red then blue. Multiply the two probabilities. After a red, there are still y blue out of x + y − 1.xx + y × yx + y − 1 = xy(x + y)(x + y − 1)
- Blue then red. After a blue, there are still x red.yx + y × xx + y − 1 = xy(x + y)(x + y − 1)
- Add the two routes. They have the same value and different routes add.xy(x + y)(x + y − 1) + xy(x + y)(x + y − 1) = 2xy(x + y)(x + y − 1)
Check. Try x = 4, y = 6: 2 × 24 ÷ (10 × 9) = 48/90 = 8/15, which matches 4/15 + 4/15 from the tree. ✓
Watch out. In a proof, show the general routes, not just a numerical example.