Chapter 12: Review

The chapter problem, solved

The warehouse fire system has these figures:

the probability of a fire on a morning is $P(F) = 0.001$ ;
the alarm sounds in a fire with probability $P(A \mid F) = 0.95$ ;
the alarm sounds with no fire with probability $P(A \mid \text{no } F) = 0.02$ .

The manager needs $P(\text{fire} \mid \text{alarm})$ .

Step 1. Write the two paths to an alarm

An alarm can happen in two ways. Multiply along each path.

P(F \text{ and } A) = P(F) \times P(A \mid F) = 0.001 \times 0.95 = 0.00095.

P(\text{no } F \text{ and } A) = P(\text{no } F) \times P(A \mid \text{no } F) = 0.999 \times 0.02 = 0.01998.

The two paths are mutually exclusive, so add them to find the probability of an alarm.

P(A) = 0.00095 + 0.01998 = 0.02093.

Step 2. Divide the fire path by the alarm total

P(F \mid A) = \frac{P(F \text{ and } A)}{P(A)} = \frac{0.00095}{0.02093} = \frac{95}{2093} \approx 0.0454.

So about $4.5\%$ of alarms are real fires. Fewer than one alarm in twenty.

Step 3. Check with a table

The same answer comes from 100000 mornings.

	Alarm sounds	Alarm does not sound	Total
Fire	95	5	100
No fire	1998	97902	99900
Total	2093	97907	100000

Of the 2093 mornings with an alarm, 95 had a fire:

\frac{95}{2093} \approx 0.0454.

There are about $21$ false alarms for each real fire, because $\dfrac{1998}{95} \approx 21$ .

Step 4. Explain the surprise

The detector is reliable. It finds 95% of fires and raises a false alarm on only 2% of quiet mornings. The surprise comes from the base rate. Fires are rare, so the 2% false-alarm rate acts on a very large number of quiet mornings. Those false alarms outnumber the fires.

This is not a fault in the detector. It is a fact about rare events, and it is why a screening test for a rare condition needs a very low false-positive rate.

Step 5. Critique the assumptions

The answer rests on three assumptions.

The mornings are independent. A faulty detector may stay triggered, and fires may cluster. Either would change the answer.
The rates are constant. The detection and false-alarm rates come from tests. Real rates change as the equipment ages.
The base rate is $0.001$ . This figure comes from records and may be uncertain. Section 12.3 showed how much the answer moves when the base rate changes.

Step 6. Answer the manager

Most alarms are false, but $P(F \mid A) \approx 0.045$ is far larger than the $0.001$ chance of a fire with no alarm. The alarm raises the risk by a factor of about 45. A real fire is serious, and a missed one is dangerous, so the manager should treat each alarm as a possible fire until it is checked.

Mixed practice — choose the method first

For each question, say which method you will use before you calculate: an addition rule, a multiplication rule, a tree diagram, the conditional probability formula, an expected count, or a method from Chapter 11. The items mix this chapter with the sampling, diagrams and summary statistics of Chapter 11.

A fair six-sided die is rolled once. Find the probability of a number greater than 4 or an odd number.
Events $A$ and $B$ are independent with $P(A) = 0.6$ and $P(B) = 0.5$ . Find $P(A \text{ and } B)$ .
For two events $A$ and $B$ , $P(A \text{ and } B) = 0.12$ and $P(B) = 0.4$ . Find $P(A \mid B)$ .
A bag has 4 white counters and 6 black counters. Two counters are taken at random, without replacement. Find the probability that the two counters are different colours.
A test for a fault has sensitivity $0.9$ and false-positive rate $0.08$ . The fault affects 5% of items. Find the probability that an item has the fault, given a positive result.
A model gives a probability of $0.04$ that a machine produces a faulty part in one hour. Find the expected number of faulty parts in 250 hours.
Find the mean of $12, 15, 16, 18, 20, 21$ .
Find the standard deviation of $12, 15, 16, 18, 20, 21$ , dividing by $n$ .
A class of width 10 contains 45 values. Find the frequency density.
A population has 450 members in one stratum and 150 in another. A stratified sample of 60 is taken. Find the number from the 450 stratum.
A line of best fit passes through the points $(2, 10)$ and $(6, 22)$ . Find its gradient.
A histogram of 200 values has frequency density 6 in a class of width 10. Estimate the probability that a value lies in this class.
A card is drawn at random from a standard pack of 52. Event $A$ is a heart. Event $B$ is a king. Find $P(A \text{ or } B)$ .
A six-sided die is rolled 300 times. A model predicts 50 of each face. The observed count of sixes is 72. Find the model prediction for sixes and the difference from the observed count.

Answers and concise solutions

1. The probability

Answer: The numbers greater than 4 are 5 and 6. The odd numbers are 1, 3 and 5. The number 5 is in both sets, so subtract it once: $\dfrac{2}{6} + \dfrac{3}{6} - \dfrac{1}{6} = \dfrac{4}{6} = \dfrac{2}{3}$ .

2. The probability of A and B

Answer: For independent events, multiply the probabilities: $0.6 \times 0.5 = 0.3 = \dfrac{3}{10}$ .

3. The conditional probability

Answer: Use the formula: $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{0.12}{0.4} = 0.3 = \dfrac{3}{10}$ .

4. The probability that the counters are different colours

Answer: The counters are different on two paths: white then black gives $\dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{24}{90}$ , and black then white gives $\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}$ . Adding gives $\dfrac{48}{90} = \dfrac{8}{15}$ .

5. The probability that the item has the fault

Answer: The fault and a positive result give $\dfrac{5}{100} \times \dfrac{9}{10} = \dfrac{45}{1000}$ . No fault and a positive result give $\dfrac{95}{100} \times \dfrac{8}{100} = \dfrac{76}{1000}$ . So $P(\text{positive}) = \dfrac{121}{1000}$ and $P(\text{fault} \mid \text{positive}) = \dfrac{45}{121} \approx 0.372$ .

6. The expected number of faulty parts

Answer: The expected number is $0.04 \times 250 = 10$ . This is a long-run average, so a particular 250 hours may give a different count.

7. The mean

Answer: The values sum to 102 over 6 values, so the mean is $\dfrac{102}{6} = 17$ .

8. The standard deviation

Answer: The mean is 17. The squared deviations sum to 56, so the variance is $\dfrac{56}{6} = \dfrac{28}{3}$ and $\sigma = \dfrac{2\sqrt{21}}{3} = 3.06$ (3 s.f.).

9. The frequency density

Answer: Frequency density is frequency divided by class width: $\dfrac{45}{10} = 4.5 = \dfrac{9}{2}$ .

10. The sample size from the 450 stratum

Answer: The population is 600, so the 450 stratum gives $\dfrac{450}{600} \times 60 = 45$ members. The other stratum gives 15, and the two add to 60.

11. The gradient

Answer: Gradient $= \dfrac{\text{change in } y}{\text{change in } x} = \dfrac{22 - 10}{6 - 2} = 3$ . The line rises by 3 for each extra unit of $x$ .

12. The estimated probability

Answer: The frequency is frequency density multiplied by class width: $6 \times 10 = 60$ . The probability is the frequency divided by the total: $\dfrac{60}{200} = \dfrac{3}{10}$ .

13. The probability

Answer: There are 13 hearts and 4 kings. The king of hearts is counted twice, so subtract it once: $\dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$ .

14. The prediction and the difference

Answer: The model predicts $\dfrac{300}{6} = 50$ sixes. The observed count is 72, which is 22 higher. The difference is large, so the model of a fair die may not fit. A refinement is to estimate the probability of a six from the observed data.

What you can now do

use the addition rule for mutually exclusive events, and the general addition rule for any two events;
test whether two events are independent, and explain why mutually exclusive events are not independent;
find a conditional probability from a two-way table, a Venn diagram or a tree diagram;
use $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$ and its rearrangement;
work backwards from a result to the probability of its cause;
build a probability model, state its assumptions, and criticise them;
refine a model and report how the answer changes;
work out an expected count and compare it with observed data;
draw on the sampling methods and summary statistics of Chapter 11 when a probability question needs data.

Your relevance reflection

Return to the prompt in the chapter opener: a test, an alarm or a check you have met that can be wrong. Write the example again. Now add three things you could not have said before this chapter: the base rate, the false-positive rate, and what a positive result actually means. Finish with one sentence about how you would judge the result the next time you meet it.