12.1 Mutually exclusive and independent events — Practise with help

Work each question closed book first. Before you calculate, decide whether the events can overlap.

Warm-up: remember

A fair six-sided die is rolled. Find the probability of rolling a 3.
Write $0.35$ as a fraction in its simplest form.
A bag has 3 red counters and 7 blue counters. One counter is taken at random. Find the probability that it is red.

Answers

One outcome out of six is a 3, so $P(3) = \dfrac{1}{6}$ .
$0.35 = \dfrac{35}{100} = \dfrac{7}{20}$ .
There are 3 red counters out of 10, so $P(\text{red}) = \dfrac{3}{10}$ .

If question 2 was hard, revise simplifying fractions before you start.

Practise with help

1. A fair six-sided die is rolled once. Find the probability of an even number or a 1.

Hint

Ask whether the two events can happen together. An even number is never 1, so the events cannot overlap.

Next step

The events are mutually exclusive, so add the two probabilities: $P(\text{even}) = \dfrac{3}{6}$ and $P(1) = \dfrac{1}{6}$ .

Full working

P(\text{even or } 1) = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}.

The events cannot both happen, so no overlap is subtracted.

2. A fair six-sided die is rolled once. Find the probability of an odd number or a prime number.

Hint

List each event. The odd numbers are $1, 3, 5$ and the primes are $2, 3, 5$ . Which numbers are in both lists?

Next step

The overlap is $\{3, 5\}$ , so $P(\text{odd and prime}) = \dfrac{2}{6}$ . Use the general addition rule and subtract this overlap once.

Full working

P(\text{odd}) = \frac{3}{6}, \qquad P(\text{prime}) = \frac{3}{6}, \qquad P(\text{odd and prime}) = \frac{2}{6}.

P(\text{odd or prime}) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3}.

The union is $\{1, 2, 3, 5\}$ , which has four outcomes, so the answer checks.

3. A fair four-sided spinner is numbered 1, 2, 3 and 4. The spinner is spun and a fair coin is tossed. Find the probability of an even number and a head.

Hint

The spinner and the coin do not affect each other, so the two results are independent.

Next step

Find $P(\text{even}) = \dfrac{2}{4}$ and $P(\text{head}) = \dfrac{1}{2}$ , then multiply them.

Full working

P(\text{even and head}) = P(\text{even}) \times P(\text{head}) = \frac{2}{4} \times \frac{1}{2} = \frac{2}{8} = \frac{1}{4}.

The events are independent, so the multiplication rule applies.

4. A survey of 120 students finds that 70 study Chemistry, 50 study Biology, and 30 study both. A student is chosen at random. Show that studying Chemistry and studying Biology are not independent.

Hint

For independence, compare $P(\text{both})$ with $P(\text{Chemistry}) \times P(\text{Biology})$ .

Next step

Work out $\dfrac{70}{120}$ , $\dfrac{50}{120}$ and $\dfrac{30}{120}$ . Multiply the first two fractions, then compare with the third.

Full working

P(\text{Chemistry}) = \frac{70}{120}, \qquad P(\text{Biology}) = \frac{50}{120}.

P(\text{Chemistry}) \times P(\text{Biology}) = \frac{70}{120} \times \frac{50}{120} = \frac{3500}{14400} = \frac{35}{144}.

The joint probability from the data is

P(\text{both}) = \frac{30}{120} = \frac{1}{4} = \frac{36}{144}.

The values $\dfrac{35}{144}$ and $\dfrac{36}{144}$ differ, so the two events are not independent.

5. In a class of 40 students, 25 walk to school, 10 cycle, and 5 both walk and cycle. A student is chosen at random. Find the probability that the student walks or cycles.

Hint

The two groups overlap. The 5 students who do both have been counted twice.

Next step

Add $\dfrac{25}{40}$ and $\dfrac{10}{40}$ , then subtract the overlap $\dfrac{5}{40}$ once.

Full working

P(\text{walk or cycle}) = \frac{25}{40} + \frac{10}{40} - \frac{5}{40} = \frac{30}{40} = \frac{3}{4}.

The 5 students who do both belong to either group, so they are counted once.

6. Two events $A$ and $B$ have $P(A) = 0.4$ and $P(B) = 0.5$ , and they are independent. Work out $P(A \text{ and } B)$ and $P(A \text{ or } B)$ . Explain why $A$ and $B$ are not mutually exclusive.

Hint

Independence gives the overlap. Mutually exclusive events have an overlap of zero.

Next step

Multiply to find $P(A \text{ and } B) = 0.4 \times 0.5$ . Then use the general addition rule with that overlap.

Full working

P(A \text{ and } B) = 0.4 \times 0.5 = 0.2.

P(A \text{ or } B) = 0.4 + 0.5 - 0.2 = 0.7.

The overlap is $0.2$ , which is not zero. The events can happen together, so they are not mutually exclusive. A non-zero overlap also rules out mutually exclusive events, because those have $P(A \text{ and } B) = 0$ .