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An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident involving a scooter driver, car driver and a truck driver are $0.01,0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. The probability that the person is a scooter driver is
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$\frac{1}{52}$
Let $\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3} \& \mathrm{~A}$ be events defined as follows. $\mathrm{E}_{1}=$ person chosen is a scooter driver $\mathrm{E}_{2}=$ person chosen is a car driver. $\mathrm{E}_{3}=$ person chosen is a truck driver $\&$ $\mathrm{A}=$ person meets with an accident $\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{2000}{12000}=\frac{1}{6} ; \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{1}{3} \& \mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{1}{2}$
$\because$ Probability that a person meets with an accident
given that he is a scooter driver $=\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)=0.01$
$\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)=0.03 \& \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{3}\right)=0.15$
- the person meets with an accident. $\therefore$ the probability that he was a scooter driver
$=\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{3}\right)}$
$=\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{1 / 6 \times 0.01}{\left(\frac{1}{6} \times 0.01\right)+\left(\frac{1}{3} \times 0.03\right)+\left(\frac{1}{2} \times 0.15\right)}=\frac{1}{52}$
$\because$ Probability that a person meets with an accident
given that he is a scooter driver $=\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)=0.01$
$\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)=0.03 \& \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{3}\right)=0.15$
- the person meets with an accident. $\therefore$ the probability that he was a scooter driver
$=\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{3}\right)}$
$=\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{1 / 6 \times 0.01}{\left(\frac{1}{6} \times 0.01\right)+\left(\frac{1}{3} \times 0.03\right)+\left(\frac{1}{2} \times 0.15\right)}=\frac{1}{52}$
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