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Exercise 9.6 · Q1
Q.

Car Repair Diagnosis. The manager of a car repair workshop knows from past experience that when a call is received from a person who is stuck up far away and has problem starting the car, the probabilities of various troubles are as follows:

EventTroubleProbability
A1A_1Battery problem0.4
A2A_2No petrol0.3
A3A_3Flooded0.1
A4A_4Some other reason0.2

Assuming that no two faults occur simultaneously and E be the event that the car starts if instructions given by the manager are followed by the driver with probabilities: P(E∣A1)=0.3P(E \mid A_1) = 0.3, P(E∣A2)=0P(E \mid A_2) = 0, P(E∣A3)=0.8P(E \mid A_3) = 0.8, P(E∣A4)=0.5P(E \mid A_4) = 0.5

  1. If a person follows the instructions given by the manager, what is the probability that the car starts?
  2. If the car starts on following the instructions of the manager, find the probability that car had a battery problem.
Yanam CbseNCERTSubjective· 5mImportance★★★★★est
95% · 20/21 Questions
✓ Free question

Total probability gives P(car starts)=0.30P(\text{car starts})=0.30; Bayes' theorem then gives P(battery problem∣starts)=0.4P(\text{battery problem}\mid\text{starts})=0.4.

Total probability: P(E)=∑iP(Ai)P(E∣Ai)P(E)=\sum_i P(A_i)P(E\mid A_i).

Bayes' theorem: P(A1∣E)=P(A1)P(E∣A1)P(E)P(A_1\mid E)=\dfrac{P(A_1)P(E\mid A_1)}{P(E)}, where A1,…,A4A_1,\dots,A_4 = trouble causes, E=E= "car starts after following instructions".

  1. Given priors and conditionals.

P(A1)=0.4, P(A2)=0.3, P(A3)=0.1, P(A4)=0.2P(A_1)=0.4,\ P(A_2)=0.3,\ P(A_3)=0.1,\ P(A_4)=0.2

P(E∣A1)=0.3, P(E∣A2)=0, P(E∣A3)=0.8, P(E∣A4)=0.5P(E\mid A_1)=0.3,\ P(E\mid A_2)=0,\ P(E\mid A_3)=0.8,\ P(E\mid A_4)=0.5

  1. Part (a): compute each joint term.

P(A1)P(E∣A1)=0.4×0.3=0.12P(A_1)P(E\mid A_1)=0.4\times0.3=0.12

P(A2)P(E∣A2)=0.3×0=0P(A_2)P(E\mid A_2)=0.3\times0=0

P(A3)P(E∣A3)=0.1×0.8=0.08P(A_3)P(E\mid A_3)=0.1\times0.8=0.08

P(A4)P(E∣A4)=0.2×0.5=0.10P(A_4)P(E\mid A_4)=0.2\times0.5=0.10

  1. Sum for total probability that the car starts.

P(E)=0.12+0+0.08+0.10=0.30P(E)=0.12+0+0.08+0.10=0.30

  1. Part (b): apply Bayes' theorem for A1A_1 (battery problem) given EE.

P(A1∣E)=P(A1)P(E∣A1)P(E)=0.120.30P(A_1\mid E)=\frac{P(A_1)P(E\mid A_1)}{P(E)}=\frac{0.12}{0.30}

  1. Simplify.

0.120.30=1230=0.4\frac{0.12}{0.30}=\frac{12}{30}=0.4

Self-check: Posterior probabilities must sum to 11: P(A1∣E)=0.4P(A_1\mid E)=0.4, P(A2∣E)=0/0.30=0P(A_2\mid E)=0/0.30=0, P(A3∣E)=0.08/0.30=0.26‾P(A_3\mid E)=0.08/0.30=0.2\overline{6}, P(A4∣E)=0.10/0.30=0.33‾P(A_4\mid E)=0.10/0.30=0.3\overline{3}; sum =0.4+0+0.267+0.333=1.000=0.4+0+0.267+0.333=1.000 ✓.

✓Final answer

  1. P(the car starts)=0.30P(\text{the car starts})=0.30.
  2. P(battery problem∣car starts)=0.4P(\text{battery problem}\mid\text{car starts})=0.4.

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