12.3 Modelling with probability and critiquing assumptions — Practise

1. The model probability

Answer: A fair spinner gives each side the same probability, so the model gives $\dfrac{1}{5} = 0.2$. The model assumes the sides are equally likely. In a real spinner the pointer may favour one side, because the sides are not identical in shape, so the equal-likelihood assumption may fail.

2. The expected number of defective components

Answer: Multiply the probability by the number of components: $0.03 \times 800 = 24$. The expected number is a long-run average, not a prediction for this exact batch.

3. The expected number of sixes

Answer: The expected number is $0.2 \times 50 = 10$. The model treats the rolls as independent and gives every roll the same probability of a 6.

4. The two conditional probabilities

Answer: (a) With a base rate of 1 in 1000, $P(\text{condition} \mid \text{positive}) = \dfrac{2}{113} \approx 0.0177$. (b) With a base rate of 5 in 100, the same test gives $\dfrac{18}{37} \approx 0.486$. The test has not changed. The base rate has changed, and it changes the answer a great deal, so a claim about a positive result is only as good as the assumed base rate.

5. The two probabilities

Answer: The independence model gives $P(\text{both fail}) = 0.02^2 = 0.0004 = \dfrac{1}{2500}$. When the supply can fail, both components fail whenever the supply fails, so $P(\text{both fail})$ is at least $0.005$. The difference is $0.005 - 0.0004 = 0.0046 = \dfrac{23}{5000}$. The shared supply makes the failures dependent, so the model is too small.

6. The expected number of wins

Answer: The expected number is $0.02 \times 500 = 10$. The model assumes each play is independent and has the same win probability.

7. 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 $\dfrac{999}{1000} \times \dfrac{1}{100} = \dfrac{999}{100000}$. So $P(\text{alarm}) = \dfrac{1094}{100000}$ and $P(\text{fire} \mid \text{alarm}) = \dfrac{95}{1094} \approx 0.0868$. Halving the false-alarm rate raises the answer from about $0.045$ to about $0.087$.

8. The predicted number

Answer: The model predicts $0.15 \times 200 = 30$ returns. The observed number is 24, so the model overestimates by 6. One refinement is to estimate the return rate from a larger, more representative sample of customers, instead of assuming $15\%$.

9. The model probability

Answer: The model gives $\dfrac{1}{12}$ for each month, because the 12 months are treated as equally likely. The assumption may fail because the number of days in a month differs and because birth rates vary through the year.

10. The probability that the item came from machine B

Answer: A defective item can come from either machine. Machine A gives $\dfrac{6}{10} \times \dfrac{2}{100} = \dfrac{12}{1000}$ defectives. Machine B gives $\dfrac{4}{10} \times \dfrac{5}{100} = \dfrac{20}{1000}$. So $P(\text{defective}) = \dfrac{32}{1000}$ and $P(\text{machine B} \mid \text{defective}) = \dfrac{20}{32} = \dfrac{5}{8}$.

11. The probability of the condition, given a positive result

Answer: The positive result can happen on two paths. Condition and positive gives $0.01 \times 0.95 = 0.0095$. Healthy and positive gives $0.99 \times 0.02 = 0.0198$. So $P(\text{positive}) = 0.0293$ and $P(\text{condition} \mid \text{positive}) = \dfrac{0.0095}{0.0293} = \dfrac{95}{293} \approx 0.324$. A positive result is more likely to come from a healthy person, because the condition is rare.

12. The total observed count and the judgement

Answer: The counts add to 600, and each count is close to the predicted 100. The model looks plausible on this evidence. A refinement would allow each face its own probability, estimated from a larger number of rolls, rather than assuming the die is fair.