Probability rules · Probability & distributions

In General Maths you found probabilities by counting, and combined two events with “and” (multiply) and “or” (add) (see probability). Further Maths uses the same rules with more events and more care about when each rule applies.

“Or”: the addition rule

If two events can happen together, adding their probabilities counts the overlap twice. Take it off once:

A or BAdd, then take off the overlap counted twice

Two special cases:

Mutually exclusive events can’t happen together, so ${P(A \cap B) = 0}$ and ${P(A \cup B) = P(A) + P(B)}$ .
Independent events don’t affect each other, so ${P(A \cap B) = P(A) \times P(B)}$ .

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q5

$X$ and $Y$ are two events such that $P(X \cup Y) = \frac{11}{15}$ and $P(X) = \frac13$ . Find $P(Y)$ if the events are:

mutually exclusive;

independent.

Mutually exclusive
- ${P(X \cup Y) = P(X) + P(Y)}$ .
- ${\frac{11}{15} = \frac13 + P(Y)}$ .
- So ${P(Y) = \frac{11}{15} - \frac{5}{15} = \frac{6}{15} = \frac25}$ .
Think first. No overlap. So P(X ∪ Y) = ?
Independent
- ${P(X \cup Y) = P(X) + P(Y) - P(X)P(Y)}$ .
- ${\frac{11}{15} = \frac13 + P(Y) - \frac13P(Y)}$ .
- Collect: ${\frac23P(Y) = \frac{11}{15} - \frac{5}{15} = \frac{6}{15}}$ .
- So ${P(Y) = \frac{6}{15} \times \frac32 = \frac35}$ .
Think first. Now the overlap is P(X) × P(Y).

More: the addition rule

Several people, each trying once

When several independent people each try once, write down each person’s chance of success and of failure. An outcome such as “only $A$ succeeds” multiplies one success and the others’ failures. An event such as “only one succeeds” adds all the outcomes that fit:

Several independent eventsPick the chances and an event

Ada (1/2)	Bola (1/3)	Chidi (1/4)	probability
✗	✗	✗	1/2 × 2/3 × 3/4 = 1/4
✗	✗	✓	1/2 × 2/3 × 1/4 = 1/12
✗	✓	✗	1/2 × 1/3 × 3/4 = 1/8
✗	✓	✓	1/2 × 1/3 × 1/4 = 1/24
✓	✗	✗	1/2 × 2/3 × 3/4 = 1/4
✓	✗	✓	1/2 × 2/3 × 1/4 = 1/12
✓	✓	✗	1/2 × 1/3 × 3/4 = 1/8
✓	✓	✓	1/2 × 1/3 × 1/4 = 1/24

11/24P(only one)

Three routes: one succeeds and the other two fail. Add them. Each row multiplies because the three are independent; the rows are added because they are different ways for the event to happen.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q13 (b)

The probabilities that Sani, Kalu and Tato will hit a target are $\frac34$ , $\frac25$ and $\frac13$ respectively. If all three men shoot once, what is the probability that the target will be hit only once?

Hits and misses
- Hits: Sani ${\frac34}$ , Kalu ${\frac25}$ , Tato ${\frac13}$ .
- Misses: Sani ${\frac14}$ , Kalu ${\frac35}$ , Tato ${\frac23}$ .
Think first. Write each man's chance of missing.
The three ways
- Only Sani: ${\frac34 \times \frac35 \times \frac23 = \frac{18}{60}}$ .
- Only Kalu: ${\frac14 \times \frac25 \times \frac23 = \frac{4}{60}}$ .
- Only Tato: ${\frac14 \times \frac35 \times \frac13 = \frac{3}{60}}$ .
Think first. Only one hits: which three outcomes?
Add
- ${\frac{18 + 4 + 3}{60} = \frac{25}{60} = \frac{5}{12}}$ .

More: several events

Sample spaces and tables

When outcomes are equally likely, count them. List the sample space (36 outcomes for two dice), or read counts from a frequency table, where the probability is the frequency divided by the total.

More: sample spaces and tables

With and without replacement

Drawing with replacement, the probabilities stay the same each time. Without replacement, both the top and the bottom drop by one after each draw. For “one of each”, remember both orders.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q15

A bag contains 4 red, 6 blue and 8 green identical marbles.

If three marbles are drawn at random, without replacement, calculate the probability that: (i) all will be green; (ii) all will have the same colour.

If each marble is replaced before another is drawn, calculate the probability that all will have the same colour.

All green, without replacement
- ${\frac{8}{18} \times \frac{7}{17} \times \frac{6}{16} = \frac{336}{4896} = \frac{7}{102}}$ .
Think first. 8 green out of 18. After one green, how many of each are left?
All the same colour
- All red: ${\frac{4 \times 3 \times 2}{4896} = \frac{24}{4896}}$ . All blue: ${\frac{6 \times 5 \times 4}{4896} = \frac{120}{4896}}$ .
- Add all three: ${\frac{24 + 120 + 336}{4896} = \frac{480}{4896} = \frac{5}{51}}$ .
Think first. Add all red, all blue and all green.
With replacement
- ${\left(\frac{4}{18}\right)^3 + \left(\frac{6}{18}\right)^3 + \left(\frac{8}{18}\right)^3 = \frac{64 + 216 + 512}{5832}}$ .
- ${= \frac{792}{5832} = \frac{11}{81}}$ .
Think first. Now each draw is out of 18 again.

More: drawing without replacement

Your turn

WAEC 2018 · Paper 2 · Q13

Probability & distributions

The probabilities that Ali, Baba and Katty will gain admission to college are $\frac23$ , $\frac34$ and $\frac45$ respectively. Find the probability that:

(a)
only Katty and Baba will gain admission;
(b)
none of them will gain admission;
(c)
at most two of them will gain admission.

Worked solution (try it first)

The chances of failing are $\frac13$ , $\frac14$ and $\frac15$ .

(a)

Only Katty and Baba: Ali fails, they succeed: $\frac13 \times \frac34 \times \frac45 = \frac15$ .

(b)

None: $\frac13 \times \frac14 \times \frac15 = \frac{1}{60}$ .

(c)

At most two is everything except all three: $1 - \frac23 \times \frac34 \times \frac45 = 1 - \frac25$
$= \frac35$ .

Report a problem with this question