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Exercise 9.5 · Q4

Q.An insurance company insures scooter drivers, car drivers and bus drivers in the ratio 4:5:3. The probability of a scooter driver, car driver and bus driver meeting with an accident is 0.7%, 0.4% and 1.2% respectively. If an insured person meets with an accident find the probability that the person is a scooter driver.

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With the 4:5:3 insurance-population split and given accident rates, Bayes' theorem gives exactly P(scooter∣accident)=13P(\text{scooter}\mid\text{accident})=\dfrac13.

Bayes' theorem with partition {S,C,B}\{S,C,B\} = scooter, car, bus drivers, and event A=A= "meets with an accident":

P(S∣A)=P(S) P(A∣S)P(S)P(A∣S)+P(C)P(A∣C)+P(B)P(A∣B)P(S\mid A)=\frac{P(S)\,P(A\mid S)}{P(S)P(A\mid S)+P(C)P(A\mid C)+P(B)P(A\mid B)}

  1. Convert the ratio 4:5:34:5:3 to probabilities (total parts =4+5+3=12=4+5+3=12):

P(S)=412,P(C)=512,P(B)=312P(S)=\frac{4}{12},\qquad P(C)=\frac{5}{12},\qquad P(B)=\frac{3}{12}

  1. Given accident probabilities:

P(A∣S)=0.007,P(A∣C)=0.004,P(A∣B)=0.012P(A\mid S)=0.007,\qquad P(A\mid C)=0.004,\qquad P(A\mid B)=0.012

  1. Compute each joint term (keep over common denominator 12):

P(S)P(A∣S)=4×0.00712=0.02812P(S)P(A\mid S)=\frac{4\times0.007}{12}=\frac{0.028}{12}

P(C)P(A∣C)=5×0.00412=0.02012P(C)P(A\mid C)=\frac{5\times0.004}{12}=\frac{0.020}{12}

P(B)P(A∣B)=3×0.01212=0.03612P(B)P(A\mid B)=\frac{3\times0.012}{12}=\frac{0.036}{12}

  1. Total probability of an accident (denominator). …

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