Question

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?

Answer 1

Hint : Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes. Probability is the state of being probable and the extent to which something is likely to happen in the particular favourable situations. Here we find the probability each event and the favourable using probability formula and the conditional probability.

Complete step-by-step answer :
Let us consider that the Event ${E_1}$ be the event where the driver is cyclist.
Similarly,
The event where the driver is the scooter driver be ${E_2}$
And the event where the driver is the car driver be ${E_3}$
Also, let us consider that the event where the person meets with an accident be the event A.
The total number of drivers is equal to sum of all the drivers that is $3000$ cyclists, $6000$ scooter drivers and $9000$ car drivers
Total number of drivers $= 3000 + 6000 + 9000 = 18000$
Now, probability of the individual events – by using the ratio of the favourable outcomes upon the total number of possible outcomes.
$P({E_1}) = P(Driver{\text{ is a cyclist)}}$
Place the given values-
$P({E_1}) = \dfrac{{3000}}{{18000}}$
Simplify the above equation –
$P({E_1}) = \dfrac{1}{6}$
$P({E_2}) = P(Driver{\text{ is a scooter driver)}}$
Place the given values-
$P({E_2}) = \dfrac{{6000}}{{18000}}$
Simplify the above equation –
$P({E_2}) = \dfrac{1}{3}$
$P({E_3}) = P(Driver{\text{ is a car driver)}}$
Place the given values-
$P({E_3}) = \dfrac{{9000}}{{18000}}$
Simplify the above equation –
$P({E_3}) = \dfrac{1}{2}$
Now, the probability of the accident involving drivers is –
$P(A|{E_1}) = P(Cyclist\,{\text{met with an accident) = 0}}{\text{.3}}$
Similarly,
$P(A|{E_2}) = P(Scooter{\text{ driver}}\,{\text{met with an accident) = 0}}{\text{.05}}$
$P(A|{E_3}) = P(Car{\text{ driver}}\,{\text{met with an accident) = 0}}{\text{.02}}$
The probability that the driver met with an accident is the cyclist by using the conditional probability formula –
$P({E_1}|A) = \dfrac{{P({E_1})P(A|{E_1})}}{{P({E_1})P(A|{E_1}) + P({E_2})P(A|{E_2}) + P({E_3})P(A|{E_3})}}$
Place the values in the above equation –
$P({E_1}|A) = \dfrac{{\left( {\dfrac{1}{6}} \right) \times 0.3}}{{\left( {\dfrac{1}{6}} \right) \times 0.3 + \left( {\dfrac{1}{3}} \right) \times 0.05 + \left( {\dfrac{1}{2}} \right) \times 0.02}}$
Convert the decimals into fractions in the above equation –
$P({E_1}|A) = \dfrac{{\left( {\dfrac{1}{6}} \right) \times \dfrac{3}{{10}}}}{{\left( {\dfrac{1}{6}} \right) \times \dfrac{3}{{10}} + \left( {\dfrac{1}{3}} \right) \times \dfrac{5}{{100}} + \left( {\dfrac{1}{2}} \right) \times \dfrac{2}{{100}}}}$
To simplify make the same denominators
$P({E_1}|A) = \dfrac{{\left( {\dfrac{1}{6}} \right) \times \dfrac{{30}}{{100}}}}{{\left( {\dfrac{1}{6}} \right) \times \dfrac{{30}}{{100}} + \left( {\dfrac{2}{6}} \right) \times \dfrac{5}{{100}} + \left( {\dfrac{3}{6}} \right) \times \dfrac{2}{{100}}}}$
When we have the same denominators simply add the numerators. Also, same denominators from the numerators denominator and the denominator denominator cancel each other.
$\Rightarrow P({E_1}|A) = \dfrac{{30}}{{30 + 10 + 6}} \\\ \Rightarrow P({E_1}|A) = \dfrac{{30}}{{46}} \;$
Take common factors from the numerator and the denominator and remove it.
$P({E_1}|A) = \dfrac{{15}}{{23}}$
Hence, the required answer is the probability that the driver is a cyclist is $\dfrac{{15}}{{23}}$
So, the correct answer is “ $\dfrac{{15}}{{23}}$ ”.

Note : Always convert the decimal numbers into fractions using tens, hundreds and simplify accordingly making the common base. Remember the probability of any event lies between zero and one.