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Exercise 9.5 · Q1
Q.

The number of loans sanctioned by a particular branch of a bank under different heads and the percentage of defaults in each category is given below:

Types of LoanNumber of Loans ApprovedDefaults (%)
Personal Loan153%
Education Loan51%
Housing Loan102%
Car Loan105%

If the loan application form picked at random for review is found to be of a person who has defaulted then find the probability that the application was for car loan.

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✓ Free question

Using Bayes' theorem over the 4 loan categories, the probability a defaulted application was a car loan is 512≈0.417\dfrac{5}{12}\approx 0.417.

Bayes' theorem with partition A1,A2,A3,A4A_1,A_2,A_3,A_4 (loan types) and event D=D= "defaulted":

P(Ai∣D)=P(Ai) P(D∣Ai)∑j=14P(Aj) P(D∣Aj)P(A_i\mid D)=\frac{P(A_i)\,P(D\mid A_i)}{\displaystyle\sum_{j=1}^{4}P(A_j)\,P(D\mid A_j)}

  1. Total loans approved: 15+5+10+10=4015+5+10+10=40.

  2. Prior probabilities (proportion of each loan type):

P(A1)=1540=0.375 (Personal),P(A2)=540=0.125 (Education)P(A_1)=\frac{15}{40}=0.375\ (\text{Personal}),\quad P(A_2)=\frac{5}{40}=0.125\ (\text{Education})

P(A3)=1040=0.25 (Housing),P(A4)=1040=0.25 (Car)P(A_3)=\frac{10}{40}=0.25\ (\text{Housing}),\quad P(A_4)=\frac{10}{40}=0.25\ (\text{Car})

  1. Given default rates: P(D∣A1)=0.03, P(D∣A2)=0.01, P(D∣A3)=0.02, P(D∣A4)=0.05P(D\mid A_1)=0.03,\ P(D\mid A_2)=0.01,\ P(D\mid A_3)=0.02,\ P(D\mid A_4)=0.05.

  2. Compute each joint term P(Ai)⋅P(D∣Ai)P(A_i)\cdot P(D\mid A_i):

0.375×0.03=0.011250.125×0.01=0.001250.375\times0.03=0.01125\qquad 0.125\times0.01=0.00125

0.25×0.02=0.005000.25×0.05=0.012500.25\times0.02=0.00500\qquad 0.25\times0.05=0.01250

  1. Total probability of default (denominator):

P(D)=0.01125+0.00125+0.00500+0.01250=0.03000P(D)=0.01125+0.00125+0.00500+0.01250=0.03000

  1. Apply Bayes' theorem for the Car loan:

P(A4∣D)=0.012500.03000=12503000=512≈0.4167P(A_4\mid D)=\frac{0.01250}{0.03000}=\frac{1250}{3000}=\frac{5}{12}\approx0.4167

Self-check: 512≈0.4167\frac{5}{12}\approx0.4167; expressing all as "defaults per 40 loans": Personal =15×0.03=0.45=15\times0.03=0.45, Education =5×0.01=0.05=5\times0.01=0.05, Housing =10×0.02=0.20=10\times0.02=0.20, Car =10×0.05=0.50=10\times0.05=0.50; total expected defaults =1.20=1.20; P(Car∣D)=0.50/1.20=5/12P(\text{Car}\mid D)=0.50/1.20=5/12 — matches.

✓Final answer

P(Car loan∣defaulted)=512≈0.417P(\text{Car loan}\mid\text{defaulted})=\dfrac{5}{12}\approx0.417 (≈ 41.7%41.7\%).

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