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Probability · Lesson 2 of 2

Combined events and tree diagrams

Two or more events together: multiply for 'and', add for 'or', tree diagrams, draws with and without replacement, and 'at least one' by the complement.

18 minYou should already know: Number foundations & fractions Sets & Venn diagrams
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  2. 2

Many questions involve two things happening: two contracts, two students passing, two balls drawn from a bag. Two rules do almost all the work:

  • “And” (one thing, then another): multiply. P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B) when the events are independent, meaning one doesn’t change the chance of the other.
  • “Or” (one way, or another way): add, as long as the ways can’t both happen.
A and BP(A) = 2⁄3P(B)= 3⁄5area = 2⁄3 × 3⁄5
Both A and BP(A and B) = P(A) × P(B): the corner of the square

Independent events

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q5

A building contractor tendered for two independent contracts, XX and YY. The probability that he will win contract XX is 0.50.5 and that he will not win contract YY is 0.30.3. What is the probability that he will win:

both contracts;

exactly one of the contracts;

neither of the contracts?

  1. Write all four probabilities

    P(X)=0.5P(X) = 0.5, so P(not X)=0.5P(\text{not } X) = 0.5.

    P(not Y)=0.3P(\text{not } Y) = 0.3, so P(Y)=1−0.3=0.7P(Y) = 1 - 0.3 = 0.7.

    Think first. He does not win Y with probability 0.3. What is P(wins Y)?

  2. (a) Both

    Wins X and wins Y: 0.5×0.7=0.350.5 \times 0.7 = 0.35.

  3. (b) Exactly one

    X but not Y, or Y but not X:

    0.5×0.3+0.5×0.7=0.15+0.35=0.50.5 \times 0.3 + 0.5 \times 0.7 = 0.15 + 0.35 = 0.5

    Think first. There are two ways to win exactly one. What are they?

  4. (c) Neither

    Not X and not Y: 0.5×0.3=0.150.5 \times 0.3 = 0.15.

    Check: the four cases (both, only X, only Y, neither) add to 0.35+0.15+0.35+0.15=10.35 + 0.15 + 0.35 + 0.15 = 1 ✓.

“At least one”: use the complement

“At least one” means one, or two, or more. That’s many cases, but the opposite is just one case: none.

P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none})
bothonly 1stonly 2ndneitherat least one = 1 − P(neither)
At least oneEverything except neither

Tree diagrams

A tree diagram shows every route. Each branch is labelled with its probability; the branches from one point add up to 1.

  • Multiply along a route to get the probability of that route.
  • Add the routes that make the event you want.
p1 − pq1 − qp × qp(1 − q)(1 − p)q(1 − p)(1 − q)
Tree diagramMultiply along a route; add the routes you want
Draw from the bag: the tree diagramDraw a ball, or change the bag and the event
RBRBRBRBBIn the bag:4 red, 5 blue1st ball2nd ball4/9R3/8RRR: 4/9 × 3/8 = 12/725/8BRB: 4/9 × 5/8 = 20/725/9B4/8RBR: 5/9 × 4/8 = 20/724/8BBB: 5/9 × 4/8 = 20/72
4/9P(same colour)12/72 + 20/72add the branches
The first ball stays out, so the second draw is out of 8, and there is one fewer of the colour already taken. Multiply along each branch; add the branches you want (keep the same denominator, 72, until the end).

Draw two balls from the bag and watch where each lands on the tree: the route lights up, and the readout multiplies along it. Then switch between with and without replacement and draw again. Without replacement the first ball stays out, so the second draw is from one fewer ball, and one fewer of the colour already taken. Change the number of red and blue balls, and pick an event to see which branches to add.

Worked example · WAEC 2021

WAEC 2021 · Paper 2 · Q5 (a)

There are 25 boys and 15 girls in a class, all of them equally likely to be chosen for a contest. If two students are chosen one after the other without replacement for the contest, find the probability that: (i) two boys are chosen; (ii) a boy and a girl are chosen; (iii) two boys or two girls are chosen.

  1. (i) Two boys

    24 boys out of 39 students:

    2540×2439=6001560=513\frac{25}{40} \times \frac{24}{39} = \frac{600}{1560} = \frac{5}{13}

    Think first. First a boy: 2540\frac{25}{40}. After that, how many boys and how many students are left?

  2. (ii) A boy and a girl

    Boy then girl, or girl then boy:

    2540×1539+1540×2539=3751560+3751560=7501560=2552\frac{25}{40} \times \frac{15}{39} + \frac{15}{40} \times \frac{25}{39} = \frac{375}{1560} + \frac{375}{1560} = \frac{750}{1560} = \frac{25}{52}

    Think first. How many routes give one of each?

  3. (iii) Two boys or two girls

    Two girls: 1540×1439=2101560\frac{15}{40} \times \frac{14}{39} = \frac{210}{1560}. Two boys was 6001560\frac{600}{1560}, so

    P=600+2101560=8101560=2752P = \frac{600 + 210}{1560} = \frac{810}{1560} = \frac{27}{52}

    Check: (ii) and (iii) together cover everything, and 2552+2752=1\frac{25}{52} + \frac{27}{52} = 1 ✓.

    Think first. What is P(two girls)?

Your turn

WAEC 2024 · Paper 2 · Q5

A basket contains 3 gold-plated marbles, 4 diamond marbles and some silver marbles, all of the same size and shape. Two marbles were drawn from the basket at random one after the other without replacement. If the probability that both marbles were silver is 115\frac{1}{15}, find the number of silver marbles.

Worked solution (try it first)
  1. Let there be ss silver marbles, so s+7s + 7 marbles in all.
  2. Without replacement, the first is silver with probability ss+7\frac{s}{s + 7}, and the second with probability s−1s+6\frac{s - 1}{s + 6} (one silver marble and one marble in total have gone).
  3. So ss+7×s−1s+6=115\frac{s}{s + 7} \times \frac{s - 1}{s + 6} = \frac{1}{15}.
  4. Cross-multiply: 15s(s−1)=(s+7)(s+6)15s(s - 1) = (s + 7)(s + 6), so 15s2−15s=s2+13s+4215s^2 - 15s = s^2 + 13s + 42.
  5. Collect terms: 14s2−28s−42=014s^2 - 28s - 42 = 0.
  6. Divide by 14: s2−2s−3=0s^2 - 2s - 3 = 0, which factorises as (s−3)(s+1)=0(s - 3)(s + 1) = 0.
  7. So s=3s = 3 or s=−1s = -1.
  8. A number of marbles can't be negative, so there are 3 silver marbles.
  9. Check: 310×29=690\frac{3}{10} \times \frac{2}{9} = \frac{6}{90}
    =115= \frac{1}{15}.

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