Elementary Probability

Lessons

Senior 1 · Mathematics

Elementary Probability

5 lessons

Lesson 5 of 5

Basic rules of probability

25 min

If the chance of rain is 35%, what is the chance of a dry day?

You already answered questions like this in Lesson 2 without a formula: either it rains or it does not, so the two chances must make up the whole. Probabilities are connected by a few simple rules. Each rule follows from counting outcomes, and together they let you find one probability from others without listing everything again.

Definition

Complement of an event: the event that $E$ does not happen. It contains every outcome of $S$ that is not in $E$ .

Notation: $E^{'}$ , read "E prime" or "not E", as for the complement of a set in Unit 1.

Example: rolling a die, if $E = {6}$ ("a six"), then $E' = \{1, 2, 3, 4, 5\}$ ("not a six").

$E$ and $E^{'}$ do not overlap, and together they fill all of $S$ .

Definition

Mutually exclusive events: events that cannot happen at the same time. As sets they have no outcome in common: $A \cap B = \varnothing$ .

Example: with one die, "a 1" and "a 6" are mutually exclusive. "An even number" and "a number greater than 3" are not, because a 4 or a 6 belongs to both.

Important

The basic rules of probability (for equally likely outcomes; the same rules hold for every probability).

$0 \le P(E) \le 1$ . Why: $0 \le n(E) \le n(S)$ ; divide by $n (S)$ .
$P (S) = 1$ and $P(\varnothing) = 0$ : a certain event has probability 1, an impossible one 0.
The probabilities of all the outcomes add up to 1. Why: each of the $n (S)$ outcomes has probability $\frac{1}{n(S)}$ , and $n(S) \times \frac{1}{n(S)} = 1$ .
Complement rule: $P (E^{'}) = 1 - P (E) .$ Why: every outcome is in exactly one of $E$ and $E^{'}$ , so $n (E) + n (E^{'}) = n (S)$ . Divide by $n (S)$ : $P (E) + P (E^{'}) = 1$ .
Addition rule for mutually exclusive events: $P(A \text{ or } B) = P(A) + P(B).$ Why: if $A$ and $B$ share no outcome, then $n(A \cup B) = n(A) + n(B)$ .

Worked example 1: the complement rule

(a) The probability of rain in Kigali tomorrow is 0.35. Find the probability that it does not rain. (b) A fair die is rolled. Find $P(\text{not a six})$ .

Solution

$P(\text{no rain}) = 1 - 0.35 = 0.65$ .
$P(\text{six}) = \frac{1}{6}$ , so $P(\text{not a six}) = 1 - \frac{1}{6} = \frac{5}{6}$ . Check by counting: $\{1, 2, 3, 4, 5\}$ has 5 of the 6 outcomes. ✓

Worked example 2: a missing probability

A spinner with unequal sectors, so its colours are not equally likely.

For this spinner, $P(\text{red}) = 0.3$ , $P(\text{blue}) = 0.25$ and $P(\text{yellow}) = 0.15$ . (a) Find $P(\text{green})$ . (b) Find $P(\text{not blue})$ . (c) Find $P(\text{red or yellow})$ .

Solution

The pointer must stop on exactly one colour, so the four probabilities add up to 1: $P(\text{green}) = 1 - (0.3 + 0.25 + 0.15) = 1 - 0.7 = 0.3$ .
$P(\text{not blue}) = 1 - 0.25 = 0.75$ .
Red and yellow cannot both happen on one spin, so $P(\text{red or yellow}) = 0.3 + 0.15 = 0.45$ .

The colours are not equally likely here, so $\frac{n(E)}{n(S)}$ cannot be used, but the rules still work.

Worked example 3: "not" and "at least one"

The 6 doubles are shaded; the other 30 outcomes are "not a double".

(a) Two dice are rolled. Find $P(\text{not a double})$ . (b) Two coins are tossed. Find $P(\text{at least one head})$ .

Solution

$P(\text{double}) = \frac{6}{36} = \frac{1}{6}$ , so $P(\text{not a double}) = 1 - \frac{1}{6} = \frac{5}{6}$ . Counting the 30 unshaded cells gives $\frac{30}{36} = \frac{5}{6}$ too, but the rule is quicker.
The complement of "at least one head" is "no heads", that is ${T T}$ , with $P = \frac{1}{4}$ . So $P(\text{at least one head}) = 1 - \frac{1}{4} = \frac{3}{4}$ .

Worked example 4: when events overlap

$A$ : an even number; $B$ : a number greater than 3. The overlap $A \cap B = \{4, 6\}$ .

A fair die is rolled. $A$ is "an even number" and $B$ is "a number greater than 3". (a) Find $P(\text{1 or 6})$ . (b) Find $P(A \text{ or } B)$ .

Solution

"1" and "6" are mutually exclusive: $P(\text{1 or 6}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$ .
$A = \{2, 4, 6\}$ and $B = \{4, 5, 6\}$ overlap in ${4, 6}$ , so they are not mutually exclusive. Adding $\frac{3}{6} + \frac{3}{6} = 1$ would say the event is certain, which is false: a 1 or a 3 is in neither. Count the union instead: $A \cup B = \{2, 4, 5, 6\}$ , so $P(A \text{ or } B) = \frac{4}{6} = \frac{2}{3}$ .

This is the Venn diagram rule of Unit 1, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ , divided by $n (S)$ :

P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6}.

Worked example 5: from probabilities back to counts

A bag holds 12 beads, some red, some blue and the rest green. A bead is taken at random. $P(\text{red}) = \frac{1}{4}$ and $P(\text{blue}) = \frac{1}{3}$ . How many beads of each colour are there, and what is $P(\text{green})$ ?

Solution Red: $\frac{1}{4} \times 12 = 3$ . Blue: $\frac{1}{3} \times 12 = 4$ . Green: $12 - 3 - 4 = 5$ , so $P(\text{green}) = \frac{5}{12}$ .

Check: $\frac{3}{12} + \frac{4}{12} + \frac{5}{12} = 1$ . ✓

Warning

Common mistakes

Adding probabilities of events that overlap. Add only when the events are mutually exclusive. If your sum is more than 1, you have certainly counted some outcomes twice.
Using $P (A) + P (B) = 1$ for any two events. That is true only when $B$ is the complement of $A$ . "A 1" and "a 6" have probabilities adding to $\frac{1}{3}$ , not 1.
Forgetting the complement in "at least one" questions, and counting every case the long way.

Tip

When a question says "not" or "at least one", think complement first: $P(\text{at least one}) = 1 - P(\text{none})$ . It is usually far quicker than counting.

The Organic Chemistry Tutor: Probability of Complementary Events and Sample Space (6 min). Watch how the probability of an event and of its complement always add up to 1.

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Practice

Give fractions in their simplest form. Questions 6 and 7 are harder.

$P (E) = 0.62$ . Find $P (E^{'})$ .
A letter is chosen at random from the cards spelling KIGALI. Find $P(\text{I})$ and $P(\text{not I})$ .
A school football team's chances in its next match are: win 0.45, draw 0.3. Find the probability that it loses.
Two dice are rolled. Find $P(\text{the total is not } 7)$ .
A card is picked at random from ten cards numbered 1 to 10. Are the events "a number less than 4" and "a number greater than 8" mutually exclusive? Find the probability that one or the other happens.
For the same ten cards, let $A$ be "an even number" and $B$ be "a multiple of 3". Find $P(A \text{ or } B)$ .
A bag holds 20 counters. $P(\text{white}) = \frac{2}{5}$ and $P(\text{black}) = \frac{1}{4}$ ; the rest are red. How many red counters are there?

Show answersHide answers

$1 - 0.62 = 0.38$
Two of the six cards show I: $P(\text{I}) = \frac{2}{6} = \frac{1}{3}$ ; $P(\text{not I}) = 1 - \frac{1}{3} = \frac{2}{3}$ .
$1 - (0.45 + 0.3) = 0.25$
$P(\text{total } 7) = \frac{6}{36} = \frac{1}{6}$ , so $P(\text{not } 7) = \frac{5}{6}$ .
Yes: no number is both less than 4 and greater than 8. $P = \frac{3}{10} + \frac{2}{10} = \frac{5}{10} = \frac{1}{2}$ .
$A = \{2, 4, 6, 8, 10\}$ , $B = \{3, 6, 9\}$ , and they overlap in ${6}$ : $P = \frac{5}{10} + \frac{3}{10} - \frac{1}{10} = \frac{7}{10}$ . Check: $A \cup B = \{2, 3, 4, 6, 8, 9, 10\}$ , 7 cards. ✓
White $\frac{2}{5} \times 20 = 8$ , black $\frac{1}{4} \times 20 = 5$ , so red $= 20 - 8 - 5 = 7$ .

1If the chance of rain is 35%, what is the chance of a dry day?

If the chance of rain is 35%, what is the chance of a dry day?