An insurance company insured 3000 cyclists, 6000 scooter drivers and 9000 car drivers. The probability of an accident involving a cyclist, a scooter driver and a car driver are 0·3, 0·05 and 0·02 respectively. One of the insured persons meets with an accident. What is the probability that he is a cyclist ?

 

 

 

 
 
 
 
 

Answers (1)

Let the events are defined as E : The person chosen is a cyclist.
E2 : Person chosen is a scooter driver.
E3 :Person chosen is a car driver and
A: The person meets with an accident.
Then P\left ( E_{1} \right )= \frac{3000}{18000}= \frac{3}{18}\: P\left ( E_{2} \right )= \frac{6000}{18000}
P\left ( E_{3} \right )= \frac{9000}{18000}= \frac{9}{18}
Also, P\left (\frac{A}{E_{1}} \right )=0\cdot 3= \frac{30}{100},P\left (\frac{A}{E_{2}} \right )=0\cdot 05= \frac{5}{100}
P\left (\frac{A}{E_{3}} \right )=0\cdot 02 = \frac{2}{100}
Thus required probability,
P\left (\frac{E_{1}} {A}\right )= \frac{P\left ( \frac{A}{E_{1}} \right )P\left ( E_{1} \right )}{P\left ( \frac{A}{E_{1}} \right )P\left ( E_{1} \right )+P\left ( \frac{A}{E_{2}} \right )P\left ( E_{2} \right )+P\left ( \frac{A}{E_{3}} \right )P\left ( E_{3} \right )}
\Rightarrow \frac{\frac{30}{100}\times \frac{3}{18}}{\frac{30}{100}\times \frac{3}{18}+\frac{5}{100}\times \frac{6}{18}+\frac{2}{100}\times \frac{9}{18}}
\Rightarrow \frac{15}{15+5+3}= \frac{15}{23}

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