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NDA Math Probability

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Question : 74 of 100

Marks: +1, -0

An insurance company insured 2000 scooter drives,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

$\frac{1}{52}$
$\frac{3}{52}$
$\frac{15}{52}$
$\frac{19}{52}$

Solution:

Let

E_1,E_2,E_3

and A be events defined as follows.

E_1

= person chosen is a scooter driver

E_2

= person chosen is a car driver.

E_3

= person chosen is a truck driver &

A = person meets with an accident

P(E_1)=\frac{2000}{12000}=\frac{1}{6};P(E_2)=\frac{1}{3}

&

P(E_3)=\frac{1}{2}

∵Probability that a person meets with an accident given that he is a scooter driver

=P(A/E_1)=0.01

P(A/E_2)=0.03\&P(A/E_3)=0.15

∵ the person meets with an accident,

∴ the probability that he was a scooter driver

=P(E_1/A)=\frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}

=P(E_1/A)=\frac{\frac{1}{6}\times0.01}{\left(\frac{1}{6}\times0.01\right)+\left(\frac{1}{3}\times0.03\right)+\left(\frac{1}{2}\times0.15\right)}=\frac{1}{52}

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