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GCSE Maths · EdexcelChapter 23 · Probability
GCSE Maths · Probability · Higher · Lesson 23.6

GCSE Maths: Tree diagrams: with and without replacement (Higher)

Part of the GCSE Maths probability chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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23.6

Tree diagrams: with and without replacement

Edexcel 1MA1: P8, P6
Grade 4-7

A tree diagram is a picture of what can happen in two or more events, one after the other. Each stage has its own set of branches. On each branch we write the probability of that result, and at the end of each complete route we write the combined outcome. Here is a bag with 3 red and 2 blue counters. One counter is picked, its colour is noted and it is put back. Then a second counter is picked.

3/5R3/5RRR = 9/252/5BRB = 6/252/5B3/5RBR = 6/252/5BBB = 4/25
Tree diagram for two picks with replacement from 3 red (R) and 2 blue (B) counters. The probabilities on the second branches are the same as on the first.

Rules for tree diagrams. (1) The probabilities on the branches leaving any one point add up to 1, because one of those things must happen. (2) To find the probability of a complete route, multiply along the branches (the “and” rule: first thing and second thing). (3) To find the probability of an event that can happen by more than one route, add the probabilities of those routes (the “or” rule: this route or that route). (4) The probabilities on all the complete routes add up to 1, which is a very good check.

Without replacement. What if the first counter is not put back? Then there are only 4 counters left for the second pick, and the numbers on the second branches change. If the first counter was red, 2 red and 2 blue are left, so the second branches are 2/4 and 2/4. If the first was blue, 3 red and 1 blue are left, so they are 3/4 and 1/4. The two picks are now dependent: the first result changes the probabilities for the second. This is the key difference: with replacement the branches are the same each time; without replacement the bottom number falls by one and the top number for the colour just taken falls by one.

3/5R2/4RRR = 3/102/4BRB = 3/102/5B3/4RBR = 3/101/4BBB = 1/10
The same bag, but the first counter is not replaced. The second-stage fractions have changed.

Not just bags. Trees work for any events: wet or dry days, a free throw that is scored or missed, a test that is passed or failed. If the probabilities on the second stage depend on the first result, draw different second branches for each first branch. And for “at least one”, remember the shortcut: find the probability of the opposite (none) and subtract from 1.

Key ideas

  • Branch probabilities from one point add up to 1.
  • Multiply along the branches (and). Add the routes you want (or).
  • The probabilities of all the complete routes add up to 1.
  • With replacement: the same branches at each stage (independent). Without replacement: the numbers change (dependent).
  • For “at least one”, use 1 − P(none).
Worked example 23.6.1
A bag has 3 red and 2 blue counters. A counter is picked, replaced, and then a second counter is picked. Find the probability that both counters are red.
3/5R3/5R2/5B2/5B3/5R2/5B

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. Draw the tree diagram. Write each probability on its branch. The branches from one point must add up to 1. The ends of the branches show every possible outcome.
    3/5R3/5RRR = 9/252/5BRB = 6/252/5B3/5RBR = 6/252/5BBB = 4/25
  2. 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
  3. 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 × 35 = 925

Check. The routes not used add up to 1625, and 925 + 1625 = 1. ✓

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

Worked example 23.6.2
Use the same bag and the same experiment (counter replaced). Find the probability that the two counters are different colours.

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: 35 × 25 = 625
    blue then red: 25 × 35 = 625
  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.
    625 + 625 = 1225

Check. The routes not used add up to 1325, and 1225 + 1325 = 1. ✓

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

Worked example 23.6.3
A bag has 4 red and 6 blue counters. Two counters are picked at random without replacement. Find the probability that they are the same colour.
4/10R3/9R6/9B6/10B4/9R5/9B

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. Draw the tree diagram. Write each probability on its branch. The branches from one point must add up to 1. The ends of the branches show every possible outcome.
    4/10R3/9RRR = 2/156/9BRB = 4/156/10B4/9RBR = 4/155/9BBB = 1/3
  2. 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
    blue then blue
  3. Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
    red then red: 410 × 39 = 1290 = 215
    blue then blue: 610 × 59 = 3090 = 13
  4. Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
    215 + 13 = 215 + 515 = 715

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

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

Worked example 23.6.4
The probability that it rains on any day is 0.3. Two days are independent. Find the probability that it rains on exactly one of the two days.
0.3R0.3R0.7D0.7D0.3R0.7D

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. Draw the tree diagram. Write each probability on its branch. The branches from one point must add up to 1. The ends of the branches show every possible outcome.
    0.3R0.3RRR = 0.090.7DRD = 0.210.7D0.3RDR = 0.210.7DDD = 0.49
  2. 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 dry
    dry then rain
  3. Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
    rain then dry: 0.3 × 0.7 = 0.21
    dry then rain: 0.7 × 0.3 = 0.21
  4. Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
    0.21 + 0.21 = 0.42

Check. The routes not used add up to 0.58, and 0.42 + 0.58 = 1. ✓

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

Worked example 23.6.5
A bag has 5 red and 3 blue counters. Two counters are picked without replacement. Find the probability that at least one is blue.

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. Use P(event) = 1 − P(opposite event). Find the routes of the opposite event on the tree, multiply along them, add them, and subtract from 1.

  1. Think about the opposite event. It is easier to find the probability that the event does not happen, then subtract from 1. The routes that do not give "at least one is blue" are:
    red then red
  2. Multiply along each of these routes. Each route is "first thing and second thing", so we multiply the branch probabilities.
    red then red: 58 × 47 = 2056 = 514
  3. Subtract from 1. The event and its opposite together make probability 1.
    1 − 514 = 914

Check. Directly: the routes that give the event add up to 914, which agrees. ✓

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

Worked example 23.6.6
A bag has 6 red and 4 blue counters. Three counters are picked without replacement. Find the probability that exactly two are red.

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 then blue
    red then blue then red
    blue then 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 then blue: 610 × 59 × 48 = 120720 = 16
    red then blue then red: 610 × 49 × 58 = 120720 = 16
    blue then red then red: 410 × 69 × 58 = 120720 = 16
  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.
    16 + 16 + 16 = 36 = 12

Check. The routes not used add up to 12, and 12 + 12 = 1. ✓

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

Worked example 23.6.7
The probability that Kim is late on Monday is 0.1. If she is late on Monday, the probability she is late on Tuesday is 0.4. If she is on time on Monday, the probability she is late on Tuesday is 0.05. Find the probability that she is late on exactly one of the two days.
0.1L0.4L0.6T0.9T0.05L0.95T

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. Draw the tree diagram. Write each probability on its branch. The branches from one point must add up to 1. The ends of the branches show every possible outcome.
    0.1L0.4LLL = 0.040.6TLT = 0.060.9T0.05LTL = 0.0450.95TTT = 0.855
  2. Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
    late then on time
    on time then late
  3. Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
    late then on time: 0.1 × 0.6 = 0.06
    on time then late: 0.9 × 0.05 = 0.045
  4. Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
    0.06 + 0.045 = 0.105

Check. The routes not used add up to 0.895, and 0.105 + 0.895 = 1. ✓

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

Watch out

With replacement the second-stage probabilities stay the same. Without replacement they change: the denominator goes down by one. Also check that your final route probabilities add up to 1, and do not add along a branch or multiply between different routes.

Exercise 23G  Tree diagrams: independent events and replacement

Non-calculator
1
FA bag has 3 red counters and 2 blue counters. A counter is picked at random, its colour is noted and it is put back. A second counter is then picked. Find the probability that...
3/5R3/5R2/5B2/5B3/5R2/5B
(a)both counters are red
(b)both counters are blue
(c)the counters are different colours
(d)at least one counter is red
2
FA bag has 5 red, 3 green and 2 blue counters. A counter is picked, replaced, and a second counter is picked. Find the probability that...
The bagRRRRRGGGBB
(a)both counters are red
(b)both counters are the same colour
(c)the first is green and the second is blue
(d)neither counter is red
3
FA fair spinner has 4 equal sections and one of them is shaded. It is spun twice. Find the probability that the shaded section comes up...
(a)both times
(b)exactly once
(c)at least once
4
PSThe probability of rain on any day is 0.4. Two days are independent. Draw a tree diagram and find the probability of...
(a)rain on both days
(b)no rain on either day
(c)rain on exactly one day
5
PSIn a game, Zara takes two free throws. The probability that she scores each throw is 0.7, independently. Find the probability that she scores...
(a)both throws
(b)exactly one throw
(c)at least one throw
6
PSTo get a licence, Rav must pass a theory test and a practical test. The probability that he passes the theory test is 0.85 and the probability that he passes the practical test is 0.9. The two tests are independent. Draw a tree diagram and find the probability that he...
(a)passes both tests
(b)passes exactly one test
(c)fails both tests
7
RA biased coin lands on heads with probability 0.6. The coin is tossed twice. Show that the probability of getting exactly one head is 0.48.
8
RMaria says: “The probabilities on the branches of a tree diagram that leave the same point always add up to 1.” Explain why she is right.
9
FA bag has 10 counters. Four are red and the rest are blue. A counter is picked at random, replaced, and another is picked. Find the probability that the counters are different colours.
10
PSA bag has 2 red, 1 blue and 1 green counter. A counter is picked, replaced, then another is picked. Find the probability that the two counters are the same colour.
11
PSA biased coin lands on heads with probability 0.6. It is tossed three times. Find the probability of getting...
(a)three heads
(b)exactly two heads
(c)at least one tail

Exercise 23H  Tree diagrams: dependent events and no replacement

Non-calculator
1
FA bag has 4 red counters and 6 blue counters. Two counters are taken at random without replacement. Find the probability that...
4/10R3/9R6/9B6/10B4/9R5/9B
(a)both counters are red
(b)both counters are blue
(c)the counters are the same colour
(d)the counters are different colours
(e)at least one counter is red
2
FA bag has 5 red, 3 green and 2 yellow counters. Two counters are taken without replacement. Find the probability that...
The bagRRRRRGGGYY
(a)both are red
(b)the first is green and the second is yellow
(c)both are the same colour
(d)neither is red
3
PSA box has 5 milk and 3 dark chocolates. Maya eats two chocolates at random. Find the probability that she eats...
(a)two dark chocolates
(b)one of each type
(c)at least one dark chocolate
4
PSA drawer holds 6 black socks and 4 white socks. Without looking, Tom takes out two socks. Find the probability that he gets a matching pair.
5
PSA class has 7 girls and 5 boys. Two students are chosen at random to give out books. Find the probability that...
(a)both are boys
(b)one is a girl and one is a boy
6
PSA raffle has 20 tickets. Jo has 5 of them. First prize and second prize are drawn at random (a ticket cannot win twice). Find the probability that Jo wins...
(a)both prizes
(b)at least one prize
7
RSam says: “A bag has 3 red and 2 blue counters. Two counters are taken without replacement. P(both red) = 35 × 35 = 925.” Explain Sam’s mistake and find the correct probability.
8
RIn a bag there are 4 red and 6 blue counters. Show that, when two counters are taken without replacement, the probability of red then blue is the same as the probability of blue then red.
9
PSA bag has 5 red and 3 blue counters. Three counters are taken without replacement. Find the probability that...
5/8R4/7R3/6R3/6B3/7B4/6R2/6B3/8B5/7R4/6R2/6B2/7B5/6R1/6B
(a)all three are red
(b)exactly two are red
(c)at least one is blue
10
PSA box contains 8 pens, of which 3 are faulty. Two pens are tested without replacement. Find the probability that at least one of them is faulty.
11
PSThe tree diagram shows the probability that Kim is late on Monday and Tuesday.
0.1L0.4L0.6T0.9T0.05L0.95T
(a)Find the probability that she is late on both days
(b)Find the probability that she is late on Tuesday
(c)Find the probability that she is late on exactly one of the two days

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