CBSE Sample Paper 2021-22 (Term 2) · Section C · 4 marks

Q14: Case study: accident-prone policyholders

Sample 2021-22 T2Ch 13 · Probability

The question

An insurance company believes that people can be divided into two classes: those who are accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at sometime within a fixed one-year period with probability 0.6, whereas this probability is 0.2 for a person who is not accident prone. The company knows that 20 percent of the population is accident prone.

Based on the given information, answer the following questions.

(i) what is the probability that a new policyholder will have an accident within a year of purchasing a policy? (2 marks)

(ii) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that he or she is accident prone? (2 marks)

Exam answer

Let $E_1$ ⁠: the policyholder is accident prone, $E_2$ ⁠: the policyholder is not accident prone, and $E$ ⁠: the new policyholder has an accident within a year of purchasing a policy.

P(E_1) = \frac{20}{100},\quad P(E_2) = \frac{80}{100}

Why?

20 percent of people are accident prone, so $P(E_1) = \frac{20}{100}$ ⁠. Everyone else is not, so $P(E_2) = 1 - \frac{20}{100} = \frac{80}{100}$ ⁠.

$E_1$ and $E_2$ do not overlap and together cover everyone. That is what the theorem of total probability needs.

P(E/E_1) = \frac{6}{10},\quad P(E/E_2) = \frac{2}{10}

Why?

The 0.6 and 0.2 in the question are conditional probabilities: the chance of an accident given the type of person.

Watch out: they are not $P(E_1)$ and $P(E_2)$ ⁠.

(i) By the theorem of total probability,

\begin{aligned} P(E) &= P(E_1)P(E/E_1) \\ &\quad + P(E_2)P(E/E_2) \\ &= \frac{20}{100} \times \frac{6}{10} + \frac{80}{100} \times \frac{2}{10} \\ &= \frac{120 + 160}{1000} = \frac{280}{1000} \end{aligned}

Why?

An accident happens either to an accident-prone person or to one who is not. Each route is a product (multiplication rule), and the two routes are added.

Arithmetic: $20 \times 6 = 120$ ⁠, $80 \times 2 = 160$ ⁠, and $100 \times 10 = 1000$ ⁠.

\begin{aligned} \frac{20}{100} \times \frac{6}{10} &= \frac{120}{1000} \\ \frac{80}{100} \times \frac{2}{10} &= \frac{160}{1000} \\[4pt] \frac{120}{1000} + \frac{160}{1000} &= \frac{280}{1000} \end{aligned}

$\therefore\ P(E) = \frac{7}{25}$

Why?

Divide the top and the bottom of $\frac{280}{1000}$ by 40: $280 \div 40 = 7$ and $1000 \div 40 = 25$ ⁠. As a decimal, $P(E) = 0.28$ ⁠.

\frac{280}{1000} = \frac{280 \div 40}{1000 \div 40} = \frac{7}{25}

(ii) By Bayes' theorem,

P(E_1/E) = \frac{P(E_1)P(E/E_1)}{P(E)}

Why?

Bayes' theorem turns the condition around: we know that $E$ happened and ask for the chance of $E_1$ ⁠.

The top is the accident-prone route from part (i). The bottom is the total $P(E)$ from part (i).

Watch out: $P(E_1/E)$ is not the same as $P(E/E_1) = 0.6$ ⁠.

= \frac{\frac{20}{100} \times \frac{6}{10}}{\frac{280}{1000}} = \frac{120}{280}

Why?

$\frac{20}{100} \times \frac{6}{10} = \frac{120}{1000}$ ⁠. Dividing by $\frac{280}{1000}$ ⁠, the 1000s cancel, leaving $\frac{120}{280}$ ⁠. Dividing the top and the bottom by 40 gives $\frac37$ ⁠.

\begin{aligned} \frac{120/1000}{280/1000} &= \frac{120}{1000} \times \frac{1000}{280} \\ &= \frac{120}{280} \\ &= \frac{120 \div 40}{280 \div 40} = \frac37 \end{aligned}

$\therefore\ P(E_1/E) = \dfrac37$

Where this comes from

Chapter 13, Probability: Total probability and Bayes' theorem.

Question type: (i) Two routes to one result; (ii) Bayes' theorem

How to spot it: click to see

(i) Type 7. Two routes to one result: total probability

Spot it: the result can happen in two (or more) ways, such as a ball from bag I or from bag II, and you want its overall chance.

Here: An accident can come from an accident-prone person (20%, then 0.6) or from someone who is not (80%, then 0.2). Multiply along each route and add: $\frac{280}{1000}$ ⁠.

(ii) Type 8. Bayes' theorem: from the result back to the cause

Spot it: the result has already happened (a red ball, a faulty item, an A grade), and you are asked for the chance that it came from one cause. It often comes as a case study or a 5-mark question.

Here: The accident has already happened, and you want the chance the person is accident prone. Divide that route, $\frac{120}{1000}$ ⁠, by the total $\frac{280}{1000}$ from (i).

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