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IGCSE / O Level Mathematics · Topic 8 Probability · Chapter 34 · pp. 488–497

Further probability

One event is easy. What about two coins, three matches, or two counters taken from a bag? This chapter lists every possible outcome in a table or a tree, then multiplies along branches and adds outcomes to get the answer.

Learning objectives

  1. Use a two-way table to list the outcomes of two events and find probabilities.
  2. Draw a tree diagram for three or more events and read probabilities from it.
  3. Multiply along branches (and) and add outcomes (or), with and without replacement.
  4. Find conditional probability P(A | B) from a table, a Venn diagram or a tree.

Book exercises: 34.1 to 34.4 and Student assessment 1 (pp. 488–497).

Combined events and two-way tables

🔑 Key ideas

  • Combined events look at two or more events together, such as two coins or two dice.
  • A two-way table lists every outcome: one event down the side, the other across the top.
  • If all outcomes are equally likely: P(event) = number of successful outcomes ÷ total number of outcomes.
  • Two coins: 2 × 2 = 4 outcomes. Two dice: 6 × 6 = 36 outcomes.
Dice 1 Dice 2 1 1 2 2 3 3 4 4 5 5 6 6 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 7 8 9 10 11 12
The 36 outcomes of two dice, each cell showing the total. The six shaded cells total 7, so P(7) = 6/36 = 1/6.

📘 Example 1 · Two coins are tossed. Find P(a head and a tail in either order)

  1. List all outcomes in a table: HH, HT, TH, TT. That is 4 equally likely outcomes.
  2. The successful outcomes are HT and TH: 2 of them.
  3. P(head and tail) = 2/4 = 1/2.
  4. Check: P(HH) = 1/4, P(TT) = 1/4, and 1/4 + 1/2 + 1/4 = 1. ✔

📘 Example 2 · Two fair tetrahedral dice (faces 1 to 4) are rolled. Find P(the sum is prime)

  1. There are 4 × 4 = 16 outcomes. The sums run from 2 to 8.
  2. The prime sums are 2, 3, 5 and 7.
  3. Count the ways: sum 2 has 1 way, sum 3 has 2, sum 5 has 4 [(1,4) (2,3) (3,2) (4,1)], sum 7 has 2 [(3,4) (4,3)].
  4. 1 + 2 + 4 + 2 = 9 successful outcomes.
  5. P(sum is prime) = 9/16.

✏️ Try it

  1. Two fair dice are rolled. Find P(a double).
  2. Two fair dice are rolled. Find P(a total of 11).
  3. Two fair dice are rolled. Find P(an even number on both dice).
Answers

1. Doubles: 6 of 36 = 1/6 · 2. (5,6) and (6,5): 2/36 = 1/18 · 3. 3 even numbers on each dice: 3 × 3 = 9, so 9/36 = 1/4

Next: tree diagrams →

Tree diagrams

🔑 Key ideas

  • A two-way table only handles two events. For three or more events use a tree diagram.
  • Each toss, roll or draw is one column. Each branch is one possible result.
  • The outcomes are listed at the right-hand ends of the branches.
  • For fair coins, every outcome is equally likely: 2 × 2 × 2 = 8 outcomes for three tosses.
Toss 1 Toss 2 Toss 3 Outcome and probability 1/2 1/2 1/2 1/2 1/2 1/2 1/2 H HHH 1/8 1/2 T HHT 1/8 1/2 H HTH 1/8 1/2 T HTT 1/8 1/2 H THH 1/8 1/2 T THT 1/8 1/2 H TTH 1/8 1/2 T TTT 1/8 H T H T H T
Three coin tosses: every branch is 1/2, and each of the 8 outcomes has probability 1/2 × 1/2 × 1/2 = 1/8.

📘 Example 1 · Three coins are tossed. Find P(exactly two heads) and P(at least one head)

  1. Read the 8 outcomes off the tree: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
  2. Exactly two heads: HHT, HTH, THH. That is 3 outcomes, so P = 3/8.
  3. At least one head means every outcome except TTT: 7 outcomes.
  4. P(at least one head) = 7/8.
  5. Check by the shortcut: 1 − P(TTT) = 1 − 1/8 = 7/8. ✔

📘 Example 2 · A computer picks the digits 1, 2 or 3 at random, twice, to make a two-digit number. Find P(even) and P(multiple of 11)

  1. Each digit has 3 choices, so there are 3 × 3 = 9 equally likely numbers: 11, 12, 13, 21, 22, 23, 31, 32, 33.
  2. Even numbers end in 2: 12, 22, 32. That is 3 of 9, so P(even) = 3/9 = 1/3.
  3. Multiples of 11: 11, 22, 33. That is 3 of 9, so P = 1/3.

✏️ Try it

  1. Three coins are tossed. Find P(all three the same).
  2. A computer picks the digits 1, 2 or 3 twice (repeats allowed). Find P(at least one 3).
  3. A cat has 3 kittens, each equally likely to be male or female. Find P(exactly one female).
Answers

1. HHH and TTT: 2/8 = 1/4 · 2. No 3 at all: 2 × 2 = 4 of 9, so 1 − 4/9 = 5/9 · 3. 3 of the 8 outcomes (FMM, MFM, MMF): 3/8

Next: multiplying branches →

Multiplying along branches

🔑 Key ideas

  • The outcomes are often not equally likely. Write each branch's probability beside it.
  • The probabilities leaving any one point add to 1.
  • The probability of an outcome is the product of the branches that lead to it.
  • With replacement: the object goes back, so the second-stage probabilities are the same as the first.
  • Without replacement: the bag changes, so the total (and one colour) goes down by 1 on the second draw.
This AND that multiply along the branches P(R,R) = 5/7 × 5/7 This OR that add the outcomes P(at least one R) = 25/49 + 10/49 + 10/49 Check all outcomes add to 1 25 + 10 + 10 + 4 = 49, so 49/49
The two tree rules, with the rain example below.

Example tree: rain on two days (P(rain) = 5/7)

StartDay 1Day 2 5/7 2/7 5/7 2/7 5/7 2/7 RNRNRN R,R 5/7 × 5/7 = 25/49R,N 5/7 × 2/7 = 10/49N,R 2/7 × 5/7 = 10/49N,N 2/7 × 2/7 = 4/49
R = rain, N = no rain. Each outcome is the product of its two branches. The four outcomes total 25 + 10 + 10 + 4 = 49 out of 49.

📘 Example 1 · Use the rain tree. Find P(rain on at least one day)

  1. The outcomes with at least one rainy day: (R,R), (R,N) and (N,R).
  2. Multiply along each: 25/49, 10/49 and 10/49.
  3. Add them: 25/49 + 10/49 + 10/49 = 45/49.
  4. Check with the shortcut: 1 − P(N,N) = 1 − 4/49 = 45/49. ✔

📘 Example 2 · A bag has 3 red and 2 blue counters. Two are taken without replacement. Find P(both red) and P(different colours)

  1. First draw: P(red) = 3/5. A red is gone, so 2 red are left among 4 counters.
  2. P(red, red) = 3/5 × 2/4 = 6/20 = 3/10.
  3. Different colours is red then blue, or blue then red.
  4. P(red, blue) = 3/5 × 2/4 = 3/10. P(blue, red) = 2/5 × 3/4 = 3/10.
  5. Add: 3/10 + 3/10 = 3/5.
  6. With replacement, P(both red) would be 3/5 × 3/5 = 9/25. Different, because the bag stays the same.

🧪 Try it: two-stage tree builder

The tree for two draws from your bag. Branch probabilities are beside each branch.

    ✏️ Try it

    1. P(rain) = 0.2 each day. Find P(rain on at least one of two days).
    2. A bag has 4 red and 1 blue counter. Two are taken without replacement. Find P(both red).
    3. Each bulb is faulty with probability 0.1. Find P(none of three bulbs faulty).
    Answers

    1. 1 − 0.8 × 0.8 = 1 − 0.64 = 0.36 · 2. 4/5 × 3/4 = 12/20 = 3/5 · 3. 0.9 × 0.9 × 0.9 = 0.729

    Next: conditional probability →

    Conditional probability

    🔑 Key ideas

    • Sometimes you are given extra information, and that changes the probability.
    • P(A | B) means "the probability of A given that B has happened" (the book prints it P(A\B)).
    • B is now the whole world: the total is the number in B, not the number overall.
    • P(A | B) = number in both A and B ÷ number in B.
    MathsArtTotal
    Boys261440
    Girls341246
    Total602686

    💡 "Given that she is a girl" means look only at the Girls row. P(Maths | girl) = 34/46 = 17/23, but P(Maths) overall = 60/86 = 30/43.

    ξ = 1 to 15 O (odd) P (prime) 1, 9, 15 3, 5, 7 11, 13 2 outside: 4, 6, 8, 10, 12, 14
    Numbers 1 to 15. O has 8 members, P has 6, and 5 are in both. P(prime) = 6/15 = 2/5, but P(prime | odd) = 5/8.

    Conditional probability on a tree (without replacement)

    StartBead 1Bead 2 10/15 5/15 9/14 5/14 10/14 4/14 RBRBRB R,R 10/15 × 9/14 = 3/7R,B 10/15 × 5/14 = 5/21B,R 5/15 × 10/14 = 5/21B,B 5/15 × 4/14 = 2/21
    10 red (R) and 5 black (B) beads, none put back. If the first bead is known to be red, ignore the dashed bottom half: P(B | first R) = 5/14.

    📘 Example 1 · Use the table. Find P(Art | boy)

    1. "Given a boy" means the total is the Boys row: 40.
    2. Boys who study Art: 14.
    3. P(Art | boy) = 14/40 = 7/20.
    4. Compare P(boy | Art) = 14/26 = 7/13. Swapping the order changes the answer.

    📘 Example 2 · 10 red and 5 black beads, none put back. Find P(second is black | first is red)

    1. The first bead is known to be red, so 9 red and 5 black are left: 14 beads.
    2. P(second black) = 5/14.
    3. Check by the tree: (10/15 × 5/14) ÷ (10/15) = 5/14. ✔

    ✏️ Try it

    1. Use the school table. Find P(boy | Maths).
    2. Use the school table. Find P(Art | girl).
    3. A bag has 3 red and 2 blue counters, none put back. The first is red. Find P(second is red).
    Answers

    1. 26/60 = 13/30 · 2. 12/46 = 6/23 · 3. 2 red left in 4 counters: 2/4 = 1/2

    Next: summary and quiz →

    Chapter summary

    • A two-way table lists the outcomes of two events. Two dice give 36 outcomes.
    • A tree diagram handles three or more events. Probabilities leaving a point add to 1.
    • And: multiply along the branches. Or: add the outcomes.
    • "At least one" is often quicker as 1 − P(none).
    • Without replacement: the total and one count go down by 1 on the next draw.
    • P(A | B) = number in A and B ÷ number in B. The total shrinks to B.

    📄 Practice questions

    The book's Exercises 34.1 to 34.4 and Student assessment 1 are in the printable assignment for this chapter.

    Mini-quiz

    1. Two fair dice are rolled. What is the probability that the total is 7?
    2. A fair coin is tossed three times. What is P(exactly one head)?
    3. A fair coin is tossed three times. What is P(at least one head)?
    4. On a tree diagram, how do you find the probability of one complete outcome?
    5. The probability of rain each day is 0.2. What is P(rain on both of two days)?
    6. A bag has 3 red and 2 blue counters. Two are taken with replacement. P(both red)?
    7. A bag has 3 red and 2 blue counters. Two are taken without replacement. P(both red)?
    8. The outcomes at the end of a complete tree diagram are all added together. The total is:
    9. What does P(A | B) mean?
    10. 46 girls are in a school; 34 of them study Maths. A girl is chosen. P(Maths | girl)?
    11. Each bulb is faulty with probability 0.1. What is P(at least one faulty in two bulbs)?
    12. A bag has 4 red and 1 blue counter. Two are taken without replacement. P(blue second)?

    Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 34 (Further probability). Explanations, the Try it answers, the tool and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.