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Questions  >>  CBSE XII  >>  Math  >>  Model Papers
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Q)

An insurance company insured 3000 scooter drivers, 5000 car drivers and 7000 truck drivers. The probabilities of their meeting with an accident respectively are 0.04, 0.05 and 0.15. One of the insured persons meets with an accident. Find the probability that he is a car driver.

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A)
Toolbox:
  • Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i/A)$ for any event A associated with $E_i$
  • Using the Bayes theorem : \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)
Step 1:
Given the total number of drivers are as follows: 3000 scooter drivers, 5000 car drivers and 7000 truck drivers.
Total = 3000+5000+7000 = 15000.
Step 2:
Let $E_1$ be the event that the insured person is a scooter driver.
P ($E_1$) = $\large\frac{\text{Number of scooter drivers}}{\text{Total number of drivers}}$
$\qquad=\large \frac{3000}{15000}$
$\qquad=\large \frac{1}{5}$
Step 3:
Let $E_2$ be the event that the insured person is a car driver.
P ($E_2$) = $\large\frac{\text{Number of car drivers}}{\text{Total number of drivers}} $
$\qquad=\large \frac{5000}{15000}$
$\qquad=\large \frac{1}{3}$
Step 4:
Let $E_3$ be the event that the insured person is a truck driver.
P ($E_3$) = $\large\frac{\text{Number of truck drivers}}{\text{Total number of drivers}}$
$\qquad=\large \frac{7000}{15000}$
$\qquad=\large \frac{7}{15}$
Step 5:
Let A: event that the insured person met with an accident.
P (scooter driver met w/ an accident) = P (A|$E_1$)
$0.04=\large\frac{4}{100}$
P (car driver met w/ an accident)= P (A|$E_2$)
$0.05=\large\frac{5}{100}$
P (truck driver met w/ an accident) = P (A|$E_3$)
$0.15=\large\frac{15}{100}$
Step 6:
The probability that a driver is a scooter driver who met w/ an accident is given by \(P(E_1/A)\).
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)+ P(E_3)P(A|E_3)}$
P ($E_1$|A) =\(\large \frac{\frac{1}{5}.\frac{4}{100}}{\frac{1}{5} .\frac{4}{100}+\frac{1}{3}.\frac{5}{100}+\frac{7}{15}.\frac{15}{100}}\)
$\Rightarrow \large\frac{4/500}{4/500+5/300+7/100}$
$\Rightarrow \large\frac{4/500}{(12+25+105)/1500}$
$\Rightarrow \large\frac{4/500}{142/1500}$
$\Rightarrow \large\frac{6}{71}$
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