12.3 Modelling with probability and critiquing assumptions — Practise with help

Work each question closed book first. Before you calculate, write down one assumption the model makes.

Warm-up: remember

For two events $A$ and $B$ , $P(A \text{ and } B) = 0.24$ and $P(B) = 0.6$ . Find $P(A \mid B)$ .
Events $A$ and $B$ are independent with $P(A) = 0.5$ and $P(B) = 0.2$ . Find $P(A \text{ or } B)$ .
Work out $15\%$ of 240.

Answers

$P(A \mid B) = \dfrac{0.24}{0.6} = 0.4 = \dfrac{2}{5}$ .
First $P(A \text{ and } B) = 0.5 \times 0.2 = 0.1$ . Then $P(A \text{ or } B) = 0.5 + 0.2 - 0.1 = 0.6 = \dfrac{3}{5}$ .
$0.15 \times 240 = 36$ .

If question 2 was hard, revise the general addition rule in Section 12.1.

Practise with help

1. A model treats a six-sided die as fair. State the probability of a 6 under the model, and give one reason the assumption may fail.

Hint

A fair die gives each face the same probability. How many faces are there?

Next step

Count the faces and give each one an equal share: $\dfrac{1}{6}$ . For the criticism, think about the shape of the die or the position of the spots.

Full working

The model gives $P(6) = \dfrac{1}{6}$ , because the six faces are assumed to be equally likely. The assumption may fail if the die is not perfectly balanced, or if the spots are drilled to different depths on different faces. A real die can favour one face.

A fair model is still useful as a starting point, because the assumption is clear and can be tested with data.

2. A model gives the probability of a fault as $0.04$ . In 350 items, find the expected number of faults.

Hint

An expected count is a probability multiplied by the number of trials.

Next step

Multiply $0.04$ by $350$ .

Full working

0.04 \times 350 = 14.

The model expects 14 faults in the batch. The expected count is a long-run average, so a single batch may differ.

3. Supplier A provides 70% of a factory parts and supplier B provides 30%. The fault rate is 1% for supplier A and 4% for supplier B. A faulty part is found. Estimate the probability that it came from supplier B.

Hint

Work out the expected number of faults from each supplier in a run of 1000 parts.

Next step

For 1000 parts, supplier A gives $0.7 \times 1000 \times 0.01 = 7$ faults, and supplier B gives $0.3 \times 1000 \times 0.04 = 12$ faults. Divide the B faults by the total faults.

Full working

\text{supplier A faults} = 0.7 \times 1000 \times 0.01 = 7,

\text{supplier B faults} = 0.3 \times 1000 \times 0.04 = 12.

P(\text{supplier B} \mid \text{faulty}) = \frac{12}{7 + 12} = \frac{12}{19}.

Supplier B provides fewer parts but more of the faults, because its fault rate is higher.

4. A test has sensitivity $0.9$ and a false-positive rate of $0.06$ . The condition affects 2% of the population. Find the probability that a person has the condition, given a positive result.

Hint

The positive result has two paths: a person with the condition, and a person without it. Write both path probabilities.

Next step

The two positive paths are $0.02 \times 0.9 = 0.018$ and $0.98 \times 0.06 = 0.0588$ . Add them, then divide the first by the total.

Full working

P(\text{positive}) = 0.02 \times 0.9 + 0.98 \times 0.06 = 0.018 + 0.0588 = 0.0768.

P(\text{condition} \mid \text{positive}) = \frac{0.018}{0.0768} = \frac{180}{768} = \frac{15}{64} \approx 0.234.

About 23% of positive results are true. The condition is uncommon, so the false positives dominate the positive results.

5. A model predicts that 25% of 400 customers will use a new service. In fact 120 do. Find the number the model predicts, the difference from the observed number, and describe one way to refine the model.

Hint

The predicted number is the model probability multiplied by 400.

Next step

Work out $0.25 \times 400 = 100$ . Subtract 100 from the observed 120.

Full working

The model predicts

0.25 \times 400 = 100

customers. The observed number is 120, which is 20 more than predicted. One refinement is to estimate the probability from a larger and more representative sample of customers, instead of using 25% from an earlier period.

6. A model assumes that two components fail independently. Each fails with probability $0.01$ . The components share a power supply that fails with probability $0.003$ , and both components fail when the supply fails. Show that the independence model underestimates the chance that both fail.

Hint

First find the chance that both fail under the independence assumption, by multiplying.

Next step

The independence model gives $0.01 \times 0.01 = 0.0001$ . Compare this with $0.003$ , the chance that the shared supply fails.

Full working

Under the independence assumption,

P(\text{both fail}) = 0.01 \times 0.01 = 0.0001.

When the shared supply fails, both components fail. The supply fails with probability $0.003$ , so the true chance that both fail is at least $0.003$ .

0.003 > 0.0001.

The shared supply makes the failures dependent, so the independence model underestimates the chance that both fail.