12.2 Conditional probability and the conditional probability formula — Practise with help

Work each question closed book first. Name the group that comes after the word "given", because that group is the denominator.

Warm-up: remember

Events $A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.6$ . Find $P(A \text{ and } B)$ .
A fair six-sided die is rolled. Find the probability of an odd number or a 2.
Write $0.45$ as a fraction in its simplest form.

Answers

Multiply the probabilities: $0.3 \times 0.6 = 0.18 = \dfrac{9}{50}$ .
The events cannot overlap, so add: $\dfrac{3}{6} + \dfrac{1}{6} = \dfrac{4}{6} = \dfrac{2}{3}$ .
$0.45 = \dfrac{45}{100} = \dfrac{9}{20}$ .

If question 1 was hard, revise the multiplication rule in Section 12.1.

Practise with help

1. A survey of 180 students finds that 90 study History, 60 study Geography, and 40 study both. A student is chosen at random from those who study History. Find the probability that the student studies Geography.

Hint

The condition is "studies History", so the 90 History students are the new total. Look at how many of them study Geography.

Next step

The overlap is 40 students, out of the 90 in the condition. Form the fraction $\dfrac{40}{90}$ .

Full working

P(\text{Geography} \mid \text{History}) = \frac{40}{90} = \frac{4}{9}.

The formula gives the same value: $\dfrac{40/180}{90/180} = \dfrac{40}{90} = \dfrac{4}{9}$ .

2. For two events $A$ and $B$ , $P(A \text{ and } B) = 0.15$ and $P(B) = 0.3$ . Find $P(A \mid B)$ .

Hint

Use $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$ .

Next step

Substitute the two values: the numerator is $0.15$ and the denominator is $0.3$ .

Full working

P(A \mid B) = \frac{0.15}{0.3} = 0.5 = \frac{1}{2}.

The condition "given $B$ " restricts the sample space to $B$ , and the overlap takes up half of it.

3. A bag contains 6 red counters and 4 green counters. Two counters are taken at random, one after the other, without replacement. Find the probability that the first counter is red and the second is green.

Hint

Draw the first two branches of a tree. The second branch is conditional on the first, so the total number of counters has dropped by one.

Next step

The first is red with probability $\dfrac{6}{10}$ . With a red counter gone, 9 counters remain and 4 are green. Multiply $\dfrac{6}{10}$ by $\dfrac{4}{9}$ .

Full working

P(\text{red then green}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}.

The counters are taken without replacement, so the two draws are not independent.

4. Use the same bag. Find the probability that the second counter is green.

Hint

The second counter is green on two paths: red then green, and green then green.

Next step

Work out each path separately. For red then green the value is $\dfrac{6}{10} \times \dfrac{4}{9}$ . For green then green it is $\dfrac{4}{10} \times \dfrac{3}{9}$ . Then add the two path probabilities.

Full working

P(\text{red then green}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90},

P(\text{green then green}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}.

P(\text{second green}) = \frac{24}{90} + \frac{12}{90} = \frac{36}{90} = \frac{2}{5}.

The two paths do not overlap, so adding them is allowed.

5. A test for a condition has sensitivity $0.8$ and a false-positive rate of $0.1$ . The condition affects 2% of the population. Find the probability that a person has the condition, given a positive result.

Hint

Draw the tree with the condition on the first branch. A positive result can happen for a person with the condition and for a person without it.

Next step

Find the two positive paths: $0.02 \times 0.8$ and $0.98 \times 0.1$ . Add them to get $P(\text{positive})$ , then divide the first path by that total.

Full working

P(\text{condition and positive}) = 0.02 \times 0.8 = 0.016,

P(\text{no condition and positive}) = 0.98 \times 0.1 = 0.098.

P(\text{positive}) = 0.016 + 0.098 = 0.114.

P(\text{condition} \mid \text{positive}) = \frac{0.016}{0.114} = \frac{16}{114} = \frac{8}{57} \approx 0.140.

The condition is rare, so most positive results come from people who do not have it.

6. For two events $A$ and $B$ , $P(A \text{ and } B) = 0.18$ , $P(B) = 0.3$ and $P(A) = 0.6$ . Find $P(A \mid B)$ and decide whether the events are independent.

Hint

Find $P(A \mid B)$ first. Then compare it with $P(A)$ .

Next step

Work out $\dfrac{0.18}{0.3}$ . Compare the result with the given $P(A) = 0.6$ .

Full working

P(A \mid B) = \frac{0.18}{0.3} = 0.6.

This equals $P(A) = 0.6$ , so knowing that $B$ happened does not change the probability of $A$ . The events are independent. The multiplication rule confirms it: $0.6 \times 0.3 = 0.18$ .