12.2 Conditional probability and the conditional probability formula — Practise

Work through these on your own. For each question, write down the group that forms the denominator before you use the formula.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A college has 200 students. 120 study Mathematics, 80 study Physics, and 50 study both. A student is chosen at random. Find the probability that the student studies Physics, given that the student studies Mathematics.
A college has 200 students. 120 study Mathematics, 80 study Physics, and 50 study both. A student is chosen at random. Find the probability that the student studies Mathematics, given that the student studies Physics.
For two events $A$ and $B$ , $P(A \text{ and } B) = 0.2$ and $P(B) = 0.5$ . Find $P(A \mid B)$ .
For two events $A$ and $B$ , $P(A \mid B) = 0.7$ and $P(B) = 0.6$ . Find $P(A \text{ and } B)$ .
For two events $A$ and $B$ , $P(A \text{ and } B) = 0.2$ , $P(B) = 0.5$ and $P(A) = 0.4$ . Find $P(A \mid B)$ and state, with a reason, whether $A$ and $B$ are independent.
A bag contains 5 red counters and 3 blue 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 blue.
A bag contains 5 red counters and 3 blue counters. Two counters are taken at random, one after the other, without replacement. Find the probability that the second counter is red.
On any morning, the probability of a fire in a warehouse is $0.001$ . A detector sounds when there is a fire with probability $0.95$ . It sounds when there is no fire with probability $0.02$ . Find the probability that there is a fire, given that the detector sounds.
In a factory, 2% of items are defective. A test detects a defective item with probability $0.9$ . It wrongly flags a good item with probability $0.05$ . Find the probability that an item is defective, given that the test flags it.
60% of email is spam. A filter flags spam with probability $0.98$ and flags genuine email with probability $0.04$ . Find the probability that a flagged email is spam, as a percentage to one decimal place.
A survey of 400 people finds that 160 have a degree and 100 work in finance. Of the people with a degree, 40 work in finance. A person is chosen at random from the 100 who work in finance. Find the probability that the person has a degree.
For two events $A$ and $B$ , $P(A \text{ and } B) = 0.21$ , $P(A) = 0.6$ and $P(B) = 0.3$ . Find $P(A \mid B)$ and $P(B \mid A)$ . Explain why the two answers are different.

Answers and concise solutions

1. The probability that the student studies Physics, given Mathematics

Answer: The condition restricts the sample space to the 120 students who study Mathematics. Of those, 50 study Physics: $P(\text{Physics} \mid \text{Mathematics}) = \dfrac{50}{120} = \dfrac{5}{12}$ .

2. The probability that the student studies Mathematics, given Physics

Answer: Now the condition restricts the sample space to the 80 students who study Physics. Of those, 50 study Mathematics: $P(\text{Mathematics} \mid \text{Physics}) = \dfrac{50}{80} = \dfrac{5}{8}$ . The two conditional probabilities are different, so the order of the condition matters.

3. The conditional probability

Answer: Use the conditional probability formula: $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{0.2}{0.5} = 0.4 = \dfrac{2}{5}$ .

4. The probability of A and B

Answer: Rearrange the formula: $P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.7 \times 0.6 = 0.42 = \dfrac{21}{50}$ .

5. The conditional probability and the test for independence

Answer: $P(A \mid B) = \dfrac{0.2}{0.5} = 0.4$ . This equals $P(A)$ , so knowing that $B$ happened does not change the probability of $A$ . The events are independent.

6. The probability of red then blue

Answer: The first counter is red with probability $\dfrac{5}{8}$ . Given that, 7 counters remain and 3 are blue, so the second is blue with probability $\dfrac{3}{7}$ . Multiply along the branch: $\dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56}$ .

7. The probability that the second counter is red

Answer: The second counter is red on two paths: red then red, and blue then red. Their probabilities are $\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56}$ and $\dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}$ . Adding the paths gives $\dfrac{35}{56} = \dfrac{5}{8}$ . The second draw has the same chance as the first.

8. The probability of a fire, given that the detector sounds

Answer: The detector sounds on two paths. Fire and alarm gives $\dfrac{1}{1000} \times \dfrac{19}{20} = \dfrac{19}{20000}$ . No fire and alarm gives $\dfrac{999}{1000} \times \dfrac{1}{50} = \dfrac{999}{50000}$ . So $P(\text{alarm}) = \dfrac{19}{20000} + \dfrac{999}{50000} = \dfrac{2093}{100000}$ . Then $P(\text{fire} \mid \text{alarm}) = \dfrac{19/20000}{2093/100000} = \dfrac{95}{2093} \approx 0.0454$ .

9. The probability that the item is defective, given that it is flagged

Answer: The flag can happen on two paths. Defective and flagged gives $\dfrac{2}{100} \times \dfrac{9}{10} = \dfrac{18}{1000}$ . Good and flagged gives $\dfrac{98}{100} \times \dfrac{5}{100} = \dfrac{490}{10000} = \dfrac{49}{1000}$ . So $P(\text{flagged}) = \dfrac{67}{1000}$ and $P(\text{defective} \mid \text{flagged}) = \dfrac{18}{67} \approx 0.269$ .

10. The probability that a flagged email is spam

Answer: Spam and flagged gives $0.6 \times 0.98 = 0.588$ . Genuine and flagged gives $0.4 \times 0.04 = 0.016$ . So $P(\text{flagged}) = 0.604$ and $P(\text{spam} \mid \text{flagged}) = \dfrac{0.588}{0.604} = \dfrac{147}{151} \approx 0.9735$ . As a percentage this is $97.4\%$ to one decimal place.

11. The probability that the person has a degree

Answer: The condition restricts the group to the 100 people who work in finance. Of those, 40 have a degree: $P(\text{degree} \mid \text{finance}) = \dfrac{40}{100} = \dfrac{2}{5}$ . The formula gives the same value, because $\dfrac{40/400}{100/400} = \dfrac{40}{100}$ .

12. The two conditional probabilities

Answer: $P(A \mid B) = \dfrac{0.21}{0.3} = \dfrac{7}{10}$ and $P(B \mid A) = \dfrac{0.21}{0.6} = \dfrac{7}{20}$ . The two conditional probabilities divide the same overlap by different totals. Dividing by a larger total gives a smaller probability, so the answers differ. In general $P(A \mid B)$ is not the same as $P(B \mid A)$ .