Board 2025 Class 12 Maths Q38 Solution: Case study: loan defaults at fixed, floating and variable rates

The question

A bank offers loan to its customers on different types of interest namely, fixed rate, floating rate and variable rate. From the past data with the bank, it is known that a customer avails loan on fixed rate, floating rate or variable rate with probabilities 10%, 20% and 70% respectively. A customer after availing loan can pay the loan or default on loan repayment. The bank data suggests that the probability that a person defaults on loan after availing it at fixed rate, floating rate and variable rate is 5%, 3% and 1% respectively.

Based on the above information, answer the following :

(i) What is the probability that a customer after availing the loan will default on the loan repayment ? (2 marks)

(ii) A customer after availing the loan, defaults on loan repayment. What is the probability that he availed the loan at a variable rate of interest ? (2 marks)

Exam answer

Let $E_1$ ⁠: the customer avails the loan on fixed rate, $E_2$ ⁠: on floating rate, $E_3$ ⁠: on variable rate, and $A$ ⁠: the customer defaults on the loan.

Why?

Each customer takes exactly one type of loan, so $E_1, E_2, E_3$ are mutually exclusive and together cover every case. This is what total probability and Bayes' theorem need. $A$ is the event we observe.

P(E_1) = \frac{1}{10},\quad P(E_2) = \frac{2}{10}

P(E_3) = \frac{7}{10}

Why?

A percentage is a fraction of $100$ ⁠: $10\% = \frac{10}{100} = \frac{1}{10}$ ⁠, and so on. Check: $\frac{1}{10} + \frac{2}{10} + \frac{7}{10} = 1$ ⁠.

P(A|E_1) = \frac{5}{100},\ P(A|E_2) = \frac{3}{100}

P(A|E_3) = \frac{1}{100}

Why?

"The probability that a person defaults after availing the loan at fixed rate" is a conditional probability: the chance of $A$ given $E_1$ ⁠. So $5\%$ is $P(A|E_1)$ ⁠, not $P(A)$ ⁠.

(i) By the theorem of total probability,

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &= \frac{1}{10}\cdot\frac{5}{100} + \frac{2}{10}\cdot\frac{3}{100} \\ &\quad + \frac{7}{10}\cdot\frac{1}{100} \\ &= \frac{5 + 6 + 7}{1000} = \frac{18}{1000} \end{aligned}

Why?

A default happens together with exactly one of the three loan types, so add the three products. Each product has denominator $10\times100 = 1000$ ⁠, with numerators $1\times5$ ⁠, $2\times3$ and $7\times1$ ⁠.

\begin{aligned} \frac{1}{10}\cdot\frac{5}{100} &= \frac{1 \times 5}{10 \times 100} = \frac{5}{1000} \\ \frac{2}{10}\cdot\frac{3}{100} &= \frac{2 \times 3}{10 \times 100} = \frac{6}{1000} \\ \frac{7}{10}\cdot\frac{1}{100} &= \frac{7 \times 1}{10 \times 100} = \frac{7}{1000} \\[4pt] 5 + 6 + 7 &= 18 \end{aligned}

$\therefore$ The required probability is $\frac{18}{1000} = \frac{9}{500}$ ⁠.

Why?

Reduce the fraction: 18 and 1000 are both even, so divide top and bottom by 2.

\frac{18}{1000} = \frac{18 \div 2}{1000 \div 2} = \frac{9}{500}

(ii) By Bayes' theorem, with $P(A)$ from (i),

\begin{aligned} P(E_3|A) &= \frac{P(E_3)P(A|E_3)}{P(A)} \\ &= \frac{\frac{7}{10}\cdot\frac{1}{100}}{\frac{18}{1000}} = \frac{\frac{7}{1000}}{\frac{18}{1000}} \end{aligned}

Why?

We know the customer defaulted and ask about the cause, so this is Bayes' theorem. Its denominator is the total probability $P(A)$ found in (i).

Both fractions have denominator $1000$ ⁠, so they cancel, leaving $\frac{7}{18}$ ⁠.

\begin{aligned} \frac{7}{10}\cdot\frac{1}{100} &= \frac{7}{1000} \\[4pt] \frac{7}{1000} \div \frac{18}{1000} &= \frac{7}{1000}\cdot\frac{1000}{18} \\ &= \frac{7}{18} \end{aligned}

Variable-rate customers default least often ( $1\%$ ⁠), but they are $70\%$ of all customers, so they make up $7$ of every $18$ defaults.

$\therefore$ The required probability is $\dfrac{7}{18}$ ⁠.

Where this comes from

Chapter 13, Probability: Case study: 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: A default can come through a fixed, floating or variable rate loan (10%, 20%, 70%). Multiply each by its default rate (5%, 3%, 1%) and add: $\frac{18}{1000}$ ⁠.

(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 customer has already defaulted, and you want the chance the loan was at a variable rate. Divide that route, $\frac{7}{1000}$ ⁠, by the total $\frac{18}{1000}$ from (i).

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Q38: Case study: loan defaults at fixed, floating and variable rates