Combined events and tree diagrams
By the end of this chapter you will be able to
- Draw a tree diagram for two selections without replacement, and find the probability of an outcome
- Use the and rule (multiply) and the or rule (add, for exclusive outcomes)
- Find the probability of at least one success by using the complement
- Use a two-way table to find a conditional probability
- Use set notation with Venn diagrams, and find expected numbers of outcomes
You should already know
- Sample spaces, two-way tables and Venn diagrams (Year 8, Chapter 12)
- Tree diagrams for independent events and expected frequency (Year 8, Chapter 12)
- Probability as a fraction and the rule that probabilities add to 1 (Year 8, Chapter 12)
Where this leads
- Probability in statistics and sampling (Year 9, Chapter 12)
- Conditional probability and independence in GCSE
- Bayes’ theorem and probability trees in GCSE
National Curriculum for England, Key Stage 3 mathematics, Year 9. Content: Use tree diagrams for dependent events, including selections without replacement, Use conditional probability with tree diagrams and two-way tables, Calculate the probability of A and B by multiplying along branches, and of A or B by adding for mutually exclusive outcomes, Calculate the probability of at least one outcome by using the complement, Use set notation for union, intersection and complement with a Venn diagram, and predict outcomes in repeated trials.
Exercise P Do you remember?
Non-calculatorMethod. There is one 6 out of six equally likely faces.
- Count the outcomes. There are 6 faces, and one of them is a 6.1 out of 6
- Write the fraction. 1/6
Check. 6 × 1/6 = 1.
Method. Name the type of events.
- Think about the meaning. If one happens, the other cannot happen.cannot both happen
- Name them. Mutually exclusive
Check. Tossing a coin gives a head or a tail, but not both.
Method. Divide the red counters by the total.
- Count. There are 5 red out of 10 counters in total.5/10
- Simplify. 5/10 = 1/2.1/2
Check. The probability of blue is also 1/2.
Method. Multiply the numerators, then the denominators.
- Multiply. 3 × 1 = 3 and 5 × 2 = 10.3/10
- Write the answer. 3/10
Check. 3/10 is 0.3, which is 3/5 of a half.
Method. List the outcomes.
- List. HH, HT, TH, TT.HH, HT, TH, TT
- Count. 4
Check. 2 × 2 = 4.
Tree diagrams for dependent events
Dependent events are events where the first outcome changes the probabilities for the second. Selecting without replacement means the counter is not put back, so both the total and the number of that colour go down by 1.
A tree diagram shows each stage as branches. Multiply the probabilities along a branch to find the probability of that whole path.
Key ideas
- Tree diagram: write the probability on every branch.
- Without replacement, both numbers in the second fraction change.
- Multiply along the branches for the probability of a whole path.
- The branches from each stage must add up to 1.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 5 counters, and 3 of them are red.P(R) = 3/5
- Write the second probability. One counter has been taken out, so there are 4 left. Take the first colour out of the bag.P(R | first R) = 1/2
- Multiply along the branches. "And" means multiply the probabilities along the branches.3/5 × 1/2 = 3/10
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 7 counters, and 3 of them are blue.P(B) = 3/7
- Write the second probability. One counter has been taken out, so there are 6 left. Take the first colour out of the bag.P(B | first B) = 1/3
- Multiply along the branches. "And" means multiply the probabilities along the branches.3/7 × 1/3 = 1/7
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 8 counters, and 5 of them are red.P(R) = 5/8
- Write the second probability. One counter has been taken out, so there are 7 left. Take the first colour out of the bag.P(B | first R) = 3/7
- Multiply along the branches. "And" means multiply the probabilities along the branches.5/8 × 3/7 = 15/56
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Watch out
Watch out: the second probability uses the new total, which is one less. Do not use the original total again.
Exam tip
Exam tip. Write the reduced numbers on the second set of branches, so you can see the change at each stage.
Exercise 11A Tree diagrams without replacement
Non-calculatorMethod. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 5 counters, and 3 of them are red.P(R) = 3/5
- Write the second probability. One counter has been taken out, so there are 4 left. Take the first colour out of the bag.P(R | first R) = 1/2
- Multiply along the branches. "And" means multiply the probabilities along the branches.3/5 × 1/2 = 3/10
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 5 counters, and 3 of them are red.P(R) = 3/5
- Write the second probability. One counter has been taken out, so there are 4 left. Take the first colour out of the bag.P(B | first R) = 1/2
- Multiply along the branches. "And" means multiply the probabilities along the branches.3/5 × 1/2 = 3/10
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 5 counters, and 2 of them are blue.P(B) = 2/5
- Write the second probability. One counter has been taken out, so there are 4 left. Take the first colour out of the bag.P(R | first B) = 3/4
- Multiply along the branches. "And" means multiply the probabilities along the branches.2/5 × 3/4 = 3/10
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 5 counters, and 2 of them are blue.P(B) = 2/5
- Write the second probability. One counter has been taken out, so there are 4 left. Take the first colour out of the bag.P(B | first B) = 1/4
- Multiply along the branches. "And" means multiply the probabilities along the branches.2/5 × 1/4 = 1/10
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 7 counters, and 3 of them are blue.P(B) = 3/7
- Write the second probability. One counter has been taken out, so there are 6 left. Take the first colour out of the bag.P(B | first B) = 1/3
- Multiply along the branches. "And" means multiply the probabilities along the branches.3/7 × 1/3 = 1/7
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 10 counters, and 5 of them are red.P(R) = 1/2
- Write the second probability. One counter has been taken out, so there are 9 left. Take the first colour out of the bag.P(B | first R) = 5/9
- Multiply along the branches. "And" means multiply the probabilities along the branches.1/2 × 5/9 = 5/18
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 10 counters, and 4 of them are red.P(R) = 2/5
- Write the second probability. One counter has been taken out, so there are 9 left. Take the first colour out of the bag.P(R | first R) = 1/3
- Multiply along the branches. "And" means multiply the probabilities along the branches.2/5 × 1/3 = 2/15
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 10 counters, and 4 of them are blue.P(B) = 2/5
- Write the second probability. One counter has been taken out, so there are 9 left. Take the first colour out of the bag.P(R | first B) = 2/3
- Multiply along the branches. "And" means multiply the probabilities along the branches.2/5 × 2/3 = 4/15
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 6 counters, and 4 of them are blue.P(B) = 2/3
- Write the second probability. One counter has been taken out, so there are 5 left. Take the first colour out of the bag.P(B | first B) = 3/5
- Multiply along the branches. "And" means multiply the probabilities along the branches.2/3 × 3/5 = 2/5
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 10 counters, and 7 of them are red.P(R) = 7/10
- Write the second probability. One counter has been taken out, so there are 9 left. Take the first colour out of the bag.P(R | first R) = 2/3
- Multiply along the branches. "And" means multiply the probabilities along the branches.7/10 × 2/3 = 7/15
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Add the probabilities of RR and BB.
- RR. P(RR) = 5/8 × 4/7 = 20/56.20/56
- BB. P(BB) = 3/8 × 2/7 = 6/56.6/56
- Add. 20/56 + 6/56 = 26/56 = 13/28.13/28
Check. The two outcomes are exclusive, so we add.
Method. Add the probabilities of RB and BR.
- RB. P(RB) = 3/5 × 2/4 = 6/20.6/20
- BR. P(BR) = 2/5 × 3/4 = 6/20.6/20
- Add. 6/20 + 6/20 = 12/20 = 3/5.3/5
Check. The four outcomes add up to 1.
Method. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 5 counters, and 3 of them are blue.P(B) = 3/5
- Write the second probability. One counter has been taken out, so there are 4 left. Take the first colour out of the bag.P(B | first B) = 1/2
- Multiply along the branches. "And" means multiply the probabilities along the branches.3/5 × 1/2 = 3/10
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Think about what has been removed.
- Look at the first branch. One counter has been taken out, so there is one fewer in the bag.total goes down by 1
- Write the answer. The number of that colour also goes down, if it was the colour taken.The total and one colour count change
Check. Dependent events change the second probabilities.
Probability of A and B, and A or B
For independent events (one does not affect the other), P(A and B) = P(A) × P(B). For mutually exclusive events, P(A or B) = P(A) + P(B).
Key ideas
- AND: multiply the probabilities (independent events).
- OR: add the probabilities (mutually exclusive events).
- Check that exclusive probabilities add to 1 or less.
- A probability of "A and B" is always smaller than either P(A) or P(B).
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 1/2 × 1/3 = 1/6
- Check. The answer is smaller than either probability.1/6 < 1/3
Check. 1/6 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Add the probabilities of the outcomes that cannot happen together.
- Identify the rule. For mutually exclusive events (they cannot both happen), "or" means add the probabilities.P(A or B) = P(A) + P(B)
- Add. 1/5 + 3/10 = 1/2
- Check. The answer must be 1 or less.1/2 ≤ 1
Check. The answer must be less than or equal to 1.
Watch out. Only add the probabilities when the events cannot happen together.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 2/5 × 1/2 = 1/5
- Check. The answer is smaller than either probability.1/5 < 2/5
Check. 1/5 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Watch out
Watch out: adding is only correct for mutually exclusive events. For two events that can both happen, you must subtract the overlap.
Exam tip
Exam tip. Ask "can both happen at the same time?" If not, use add for "or". If they are independent, use multiply for "and".
Exercise 11B The and rule and the or rule
Non-calculatorMethod. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 1/2 × 1/3 = 1/6
- Check. The answer is smaller than either probability.1/6 < 1/3
Check. 1/6 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 3/4 × 2/5 = 3/10
- Check. The answer is smaller than either probability.3/10 < 2/5
Check. 3/10 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 2/3 × 1/4 = 1/6
- Check. The answer is smaller than either probability.1/6 < 1/4
Check. 1/6 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 1/6 × 1/6 = 1/36
- Check. The answer is smaller than either probability.1/36 < 1/6
Check. 1/36 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 4/5 × 1/2 = 2/5
- Check. The answer is smaller than either probability.2/5 < 1/2
Check. 2/5 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 3/10 × 7/10 = 21/100
- Check. The answer is smaller than either probability.21/100 < 3/10
Check. 21/100 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Add the probabilities of the outcomes that cannot happen together.
- Identify the rule. For mutually exclusive events (they cannot both happen), "or" means add the probabilities.P(A or B) = P(A) + P(B)
- Add. 1/5 + 3/10 = 1/2
- Check. The answer must be 1 or less.1/2 ≤ 1
Check. The answer must be less than or equal to 1.
Watch out. Only add the probabilities when the events cannot happen together.
Method. Add the probabilities of the outcomes that cannot happen together.
- Identify the rule. For mutually exclusive events (they cannot both happen), "or" means add the probabilities.P(A or B) = P(A) + P(B)
- Add. 1/4 + 1/3 = 7/12
- Check. The answer must be 1 or less.7/12 ≤ 1
Check. The answer must be less than or equal to 1.
Watch out. Only add the probabilities when the events cannot happen together.
Method. Add the probabilities of the outcomes that cannot happen together.
- Identify the rule. For mutually exclusive events (they cannot both happen), "or" means add the probabilities.P(A or B) = P(A) + P(B)
- Add. 2/7 + 1/7 = 3/7
- Check. The answer must be 1 or less.3/7 ≤ 1
Check. The answer must be less than or equal to 1.
Watch out. Only add the probabilities when the events cannot happen together.
Method. Add the probabilities of the outcomes that cannot happen together.
- Identify the rule. For mutually exclusive events (they cannot both happen), "or" means add the probabilities.P(A or B) = P(A) + P(B)
- Add. 1/2 + 1/8 = 5/8
- Check. The answer must be 1 or less.5/8 ≤ 1
Check. The answer must be less than or equal to 1.
Watch out. Only add the probabilities when the events cannot happen together.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 1/4 × 1/2 = 1/8
- Check. The answer is smaller than either probability.1/8 < 1/4
Check. 1/8 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Add the probabilities of the outcomes that cannot happen together.
- Identify the rule. For mutually exclusive events (they cannot both happen), "or" means add the probabilities.P(A or B) = P(A) + P(B)
- Add. 1/6 + 1/6 = 1/3
- Check. The answer must be 1 or less.1/3 ≤ 1
Check. The answer must be less than or equal to 1.
Watch out. Only add the probabilities when the events cannot happen together.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 2/5 × 1/3 = 2/15
- Check. The answer is smaller than either probability.2/15 < 1/3
Check. 2/15 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Think about the overlap.
- Count the overlap. The outcomes in both A and B are counted in P(A) and again in P(B).overlap counted twice
- Write the answer. So subtract P(A and B) once.P(A or B) = P(A) + P(B) − P(A and B)
Check. This is the general addition rule.
At least one: using the complement
The complement of an event is the event not happening. P(at least one) = 1 − P(none). To find P(none), multiply the probabilities of each event not happening.
Key ideas
- P(at least one) = 1 − P(none).
- P(none) is the product of the probabilities of each event failing.
- "At least one" usually means the complement is quicker.
- Check that your answer is between 0 and 1.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 1/2) × (1 − 1/2) = 1/2 × 1/2 = 1/4
- Subtract from 1. 1 − 1/4 = 3/4
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 1/3) × (1 − 1/4) = 2/3 × 3/4 = 1/2
- Subtract from 1. 1 − 1/2 = 1/2
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 2/5) × (1 − 1/2) = 3/5 × 1/2 = 3/10
- Subtract from 1. 1 − 3/10 = 7/10
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Watch out
Watch out: do not add the two probabilities for "at least one". 1/3 + 1/4 counts the case where both happen twice.
Exam tip
Exam tip. If a question says "at least one", think about the complement first: "neither" or "none".
Exercise 11C At least one (the complement)
Non-calculatorMethod. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 1/2) × (1 − 1/2) = 1/2 × 1/2 = 1/4
- Subtract from 1. 1 − 1/4 = 3/4
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 1/3) × (1 − 1/4) = 2/3 × 3/4 = 1/2
- Subtract from 1. 1 − 1/2 = 1/2
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 2/5) × (1 − 1/5) = 3/5 × 4/5 = 12/25
- Subtract from 1. 1 − 12/25 = 13/25
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 3/4) × (1 − 1/2) = 1/4 × 1/2 = 1/8
- Subtract from 1. 1 − 1/8 = 7/8
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 1/6) × (1 − 1/6) = 5/6 × 5/6 = 25/36
- Subtract from 1. 1 − 25/36 = 11/36
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 4/5) × (1 − 1/10) = 1/5 × 9/10 = 9/50
- Subtract from 1. 1 − 9/50 = 41/50
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 2/5) × (1 − 1/3) = 3/5 × 2/3 = 2/5
- Subtract from 1. 1 − 2/5 = 3/5
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 1/2) × (1 − 1/2) = 1/2 × 1/2 = 1/4
- Subtract from 1. 1 − 1/4 = 3/4
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Think about the overlap.
- Add the two chances. Each toss gives a head with probability 1/2, so 1/2 + 1/2 = 1.1
- Check the overlap. Adding counts HH twice, and 1 is wrong.the answer is 3/4
- Write the answer. Using the complement, P(none) = 1/4, so the answer is 3/4.3/4
Check. The four outcomes are HH, HT, TH and TT. Three have at least one head.
Conditional probability with two-way tables and tree diagrams
A conditional probability P(A given B) is the probability of A when we only look at the outcomes in B. Use a two-way table or a tree diagram, and count the outcomes in the group given.
Key ideas
- P(A given B) = number in A and B ÷ number in B.
- The group given is the new sample space.
- In a tree diagram, use the branch for the given event.
- Do not use the total when the question gives a group.
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 30 boys.total in group = 30
- Count the event in that group. Count how many of those are also cycles.number = 12
- Write the probability. 2/5
Check. 12 out of 30 is 2/5.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 20 pupils who cycle.total in group = 20
- Count the event in that group. Count how many of those are also a boy.number = 12
- Write the probability. 3/5
Check. 12 out of 20 is 3/5.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 20 girls.total in group = 20
- Count the event in that group. Count how many of those are also plays chess.number = 8
- Write the probability. 2/5
Check. 8 out of 20 is 2/5.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
Watch out
Watch out: P(boy given cycles) and P(cycles given boy) are different probabilities. Use the group in the condition each time.
Exam tip
Exam tip. Write the group in the condition on your working, so the denominator is the right total.
Exercise 11D Conditional probability
Non-calculator| Cycles | Does not cycle | Total | ||
|---|---|---|---|---|
| Boys | 12 | 18 | 30 | |
| Girls | 8 | 12 | 20 | |
| Total | 20 | 30 | 50 |
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 30 boys.total in group = 30
- Count the event in that group. Count how many of those are also cycles.number = 12
- Write the probability. 2/5
Check. 12 out of 30 is 2/5.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
| Cycles | Does not cycle | Total | ||
|---|---|---|---|---|
| Boys | 12 | 18 | 30 | |
| Girls | 8 | 12 | 20 | |
| Total | 20 | 30 | 50 |
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 20 pupils who cycle.total in group = 20
- Count the event in that group. Count how many of those are also a boy.number = 12
- Write the probability. 3/5
Check. 12 out of 20 is 3/5.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
| Cycles | Does not cycle | Total | ||
|---|---|---|---|---|
| Boys | 12 | 18 | 30 | |
| Girls | 8 | 12 | 20 | |
| Total | 20 | 30 | 50 |
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 30 pupils who do not cycle.total in group = 30
- Count the event in that group. Count how many of those are also a girl.number = 12
- Write the probability. 2/5
Check. 12 out of 30 is 2/5.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 24 students who like football.total in group = 24
- Count the event in that group. Count how many of those are also a boy.number = 15
- Write the probability. 5/8
Check. 15 out of 24 is 5/8.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
Method. Use the counters left after one red is removed.
- Counters left. 3 red and 6 blue are left, so 9 in total.9
- Blue probability. P(blue) = 6/9.6/9
- Simplify. 6/9 = 2/3.2/3
Check. This is conditional on the first counter being red.
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 36 adults.total in group = 36
- Count the event in that group. Count how many of those are also owns a phone.number = 27
- Write the probability. 3/4
Check. 27 out of 36 is 3/4.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
Method. Compare the two groups.
- Whole group. P(cycles) uses all 50 pupils.20/50
- Boys only. P(cycles given boy) uses the 30 boys only.12/30
- Write the answer. The group changes, so the probability can change.The group in the condition is different
Check. 20/50 = 2/5, but 12/30 = 2/5 in this case, so they can be the same or different.
Set notation, Venn diagrams and expected frequency
The union A ∪ B means "in A or B or both". The intersection A ∩ B means "in both A and B". The complement A′ means "not in A".
The expected frequency of an outcome in n trials is n × P(outcome).
Key ideas
- A ∪ B: in A or B (or both).
- A ∩ B: in both A and B.
- n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Expected number = probability × number of trials.
Method. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 18, n(B) = 15, n(A ∩ B) = 8
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 18 + 15 − 8 = 25
- Write the answer. 25
Check. Neither: 30 − 25 = 5.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 3/10 × 50 = 15
- Check. The expected number cannot be more than the number of trials.15 ≤ 50
Check. About 15 out of 50 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 1/4 × 60 = 15
- Check. The expected number cannot be more than the number of trials.15 ≤ 60
Check. About 15 out of 60 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Watch out
Watch out: the overlap is in both sets, so it is counted twice when you add. Subtract it once.
Exam tip
Exam tip. Fill the centre of a Venn diagram first, then work outwards to the other regions.
Exercise 11E Venn diagrams and expected frequency
Non-calculatorMethod. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 18, n(B) = 15, n(A ∩ B) = 8
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 18 + 15 − 8 = 25
- Write the answer. 25
Check. Neither: 30 − 25 = 5.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 12, n(B) = 10, n(A ∩ B) = 4
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 12 + 10 − 4 = 18
- Write the answer. 18
Check. Neither: 25 − 18 = 7.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 20, n(B) = 14, n(A ∩ B) = 9
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 20 + 14 − 9 = 25
- Write the answer. 25
Check. Neither: 40 − 25 = 15.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 15, n(B) = 9, n(A ∩ B) = 3
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 15 + 9 − 3 = 21
- Write the answer. 21
Check. Neither: 30 − 21 = 9.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 25, n(B) = 18, n(A ∩ B) = 10
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 25 + 18 − 10 = 33
- Write the answer. 33
Check. Neither: 50 − 33 = 17.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Subtract the number in A or B from the total.
- Number in A or B. 18 + 15 − 8 = 25.25
- Subtract from 30. 30 − 25 = 5.5
- Write the answer. 5
Check. 5 + 25 = 30.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 1/5 × 60 = 12
- Check. The expected number cannot be more than the number of trials.12 ≤ 60
Check. About 12 out of 60 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 1/6 × 120 = 20
- Check. The expected number cannot be more than the number of trials.20 ≤ 120
Check. About 20 out of 120 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 1/4 × 200 = 50
- Check. The expected number cannot be more than the number of trials.50 ≤ 200
Check. About 50 out of 200 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Method. The complement is the probability of not happening.
- Use the complement rule. P(A′) = 1 − P(A).1 − 2/5
- Work out. 1 = 5/5, so 5/5 − 2/5 = 3/5.3/5
- Write the answer. 3/5
Check. 2/5 + 3/5 = 1.
Method. Describe each region.
- Union. A ∪ B covers both circles, including the overlap.A ∪ B = everything in A or B
- Intersection. A ∩ B is only the overlapping middle part.A ∩ B = the overlap
- Write the answer. Union is everything; intersection is only the overlap
Check. The intersection is always inside the union.
Method. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 32, n(B) = 25, n(A ∩ B) = 15
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 32 + 25 − 15 = 42
- Write the answer. 42
Check. Neither: 50 − 42 = 8.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 3/10 × 200 = 60
- Check. The expected number cannot be more than the number of trials.60 ≤ 200
Check. About 60 out of 200 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Chapter summary
| Skill | Method |
|---|---|
| Tree diagram | Probabilities on branches. Multiply along a path. Without replacement, the second probabilities change. |
| And | Multiply probabilities for independent events. |
| Or | Add probabilities for mutually exclusive events. Subtract the overlap otherwise. |
| At least one | 1 − P(none). |
| Conditional probability | Use only the group in the condition as the denominator. |
| Venn and expected frequency | n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Expected = probability × trials. |
Can you…?
- I can draw a tree diagram for dependent events and find a path probability
- I can use the and rule and the or rule correctly
- I can find the probability of at least one outcome using the complement
- I can find a conditional probability from a two-way table
- I can use set notation with Venn diagrams, and find expected numbers
Key terms
- dependent events
- events where the first outcome changes the probability of the second
- mutually exclusive
- events that cannot both happen at the same time
- complement
- the event that the given event does not happen
- union
- A ∪ B: the outcomes in A or B or both
- intersection
- A ∩ B: the outcomes in both A and B
- expected frequency
- the number of times an outcome is expected to happen in a number of trials
Exercise 11Z Check-up
Non-calculatorMethod. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 8 counters, and 5 of them are red.P(R) = 5/8
- Write the second probability. One counter has been taken out, so there are 7 left. Take the first colour out of the bag.P(R | first R) = 4/7
- Multiply along the branches. "And" means multiply the probabilities along the branches.5/8 × 4/7 = 5/14
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 1/4 × 1/5 = 1/20
- Check. The answer is smaller than either probability.1/20 < 1/5
Check. 1/20 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Add the probabilities of the outcomes that cannot happen together.
- Identify the rule. For mutually exclusive events (they cannot both happen), "or" means add the probabilities.P(A or B) = P(A) + P(B)
- Add. 1/3 + 1/6 = 1/2
- Check. The answer must be 1 or less.1/2 ≤ 1
Check. The answer must be less than or equal to 1.
Watch out. Only add the probabilities when the events cannot happen together.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 2/3) × (1 − 1/4) = 1/3 × 3/4 = 1/4
- Subtract from 1. 1 − 1/4 = 3/4
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. The condition shrinks the sample space to the group given.
- Identify the group. Only look at the people in the group: the 18 pupils who like maths.total in group = 18
- Count the event in that group. Count how many of those are also a girl.number = 12
- Write the probability. 2/3
Check. 12 out of 18 is 2/3.
Watch out. Conditional probability: use only the group in the condition, not the whole table.
Method. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 22, n(B) = 18, n(A ∩ B) = 10
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 22 + 18 − 10 = 30
- Write the answer. 30
Check. Neither: 40 − 30 = 10.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 1/2 × 30 = 15
- Check. The expected number cannot be more than the number of trials.15 ≤ 30
Check. About 15 out of 30 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Method. Add P(BB) and P(GG).
- BB. 4/8 × 3/7 = 12/56.12/56
- GG. 4/8 × 3/7 = 12/56.12/56
- Add. 12/56 + 12/56 = 24/56 = 3/7.3/7
Check. The two outcomes are mutually exclusive.
Method. Use the complement.
- Complement rule. 1 − 3/8 = 8/8 − 3/8.5/8
- Write the answer. 5/8
Check. 3/8 + 5/8 = 1.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 1/6 × 1/6 = 1/36
- Check. The answer is smaller than either probability.1/36 < 1/6
Check. 1/36 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Total 19 marks. Allow about 19 minutes.
Exercise 11X Practice test questions
Non-calculatorMethod. Multiply along the branches of the tree. The second probability depends on the first, so use the reduced numbers.
- Write the first probability. There are 10 counters, and 6 of them are blue.P(B) = 3/5
- Write the second probability. One counter has been taken out, so there are 9 left. Take the first colour out of the bag.P(B | first B) = 5/9
- Multiply along the branches. "And" means multiply the probabilities along the branches.3/5 × 5/9 = 1/3
Check. The probabilities of all four outcomes add up to 1.
Watch out. Without replacement, the total and the number of the colour taken out both go down by 1 for the second draw.
Method. Multiply the probabilities for events that do not affect each other.
- Identify the rule. For independent events, "and" means multiply the probabilities.P(A and B) = P(A) × P(B)
- Multiply. 2/5 × 1/4 = 1/10
- Check. The answer is smaller than either probability.1/10 < 1/4
Check. 1/10 is smaller than each separate probability, as expected.
Watch out. "And" means multiply. "Or" means add, but only for outcomes that cannot happen together.
Method. Find the probability of neither event happening, then subtract it from 1.
- Use the complement. "At least one" is the opposite of "neither", so find P(neither) and subtract from 1.P(at least one) = 1 − P(neither)
- Find P(neither). Both events fail: multiply the probabilities of failing.P(neither) = (1 − 1/4) × (1 − 1/5) = 3/4 × 4/5 = 3/5
- Subtract from 1. 1 − 3/5 = 2/5
Check. Adding up the outcomes with at least one success gives the same answer.
Watch out. For "at least one" it is quicker to use the complement than to add up every case.
Method. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Write the numbers. A is the set in A, B is the set in B, and both sets overlap in the middle.n(A) = 28, n(B) = 20, n(A ∩ B) = 12
- Add the two sets. Adding counts the middle part twice, so take it away once.n(A ∪ B) = 28 + 20 − 12 = 36
- Write the answer. 36
Check. Neither: 50 − 36 = 14.
Watch out. The overlap is counted in both sets, so subtract it once.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 1/2 × 80 = 40
- Check. The expected number cannot be more than the number of trials.40 ≤ 80
Check. About 40 out of 80 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Method. Use the reduced bag.
- Counters left. 3 red and 4 green are left, so 7 in total.7
- Red probability. P(red) = 3/7.3/7
- Write the answer. 3/7
Check. Conditional on the first counter being green.
Method. Find the union first.
- Union. 16 + 14 − 6 = 24.24
- Neither. 40 − 24 = 16.16
- Write the answer. 16
Check. 24 + 16 = 40.
Method. Use the complement.
- Complement. 1 − 0.35 = 0.65.0.65
- Write the answer. 0.65
Check. 0.35 + 0.65 = 1.
Method. Multiply the probability by the number of trials to get the expected number of outcomes.
- Write the rule. Expected number = probability × number of trials.expected = p × n
- Multiply. 1/2 × 150 = 75
- Check. The expected number cannot be more than the number of trials.75 ≤ 150
Check. About 75 out of 150 trials.
Watch out. The expected number is an average, so it may not be a whole number.
Method. Look at the overlap.
- Count. The overlap is counted in both P(A) and P(B).overlap counted twice
- Write the answer. Subtract P(A and B) once.Subtract the overlap
Check. P(A or B) = P(A) + P(B) − P(A and B).