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  • 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 pe

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 E1  : 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}

Posted by

Ravindra Pindel

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