Question

A car hire firm has 2 cars which it hires out day by day. If the number of demands for a car on each day follows poisson distribution with parameter 1.5, then the probability that some demand is refused is?
\[
  {\text{A}}{\text{. 1}}{\text{.12 }} \times {\text{ }}{{\text{e}}^{ - 1.5}} \\
  {\text{B}}{\text{. 1}}{\text{.25 }} \times {\text{ }}{{\text{e}}^{ - 1.5}} \\
  {\text{C}}{\text{. 1 - 3}}{\text{.625 }} \times {\text{ }}{{\text{e}}^{ - 1.5}} \\
  {\text{D}}{\text{. 3}}{\text{.625 }} \times {\text{ }}{{\text{e}}^{ - 1.5}} \\
\]

Answer 1

Hint: In order to find the probability that some demand is refused, we use the concept of poisson distribution with the help of given data. We use the formula of poisson distribution. The proportion of days some demand is refused is nothing but the probability of demand is more than 2 cars per day.

Complete step-by-step answer:
Given Data,
The hire has 2 cars.
The parameter of the poisson distribution, i.e. the mean of the distribution is equal to 2.

We know the formula of Poisson distribution is given by $ {\text{P}}\left( {{\text{x; }}\mu } \right) = \dfrac{{\left( {{{\text{e}}^{ - \mu }}} \right)\left( {{\mu ^{\text{x}}}} \right)}}{{{\text{x!}}}} $ , where x is the random variable of the distribution, e is a constant and μ is the mean of the distribution.

Let X be a random variable which represents the demands of a car on any day.
We are supposed to find the probability X > 2. The demand is reduced only when the demand exceeds the supply the hire firm can provide, hence X > 2.
So given the variable X is poisson distributed with a parameter 1.5, i.e. μ = 1.5.

Therefore the formula probability of P (X = i) is given by
 $ {\text{P}}\left( {{\text{X = i}}} \right) = \dfrac{{\left( {{{\text{e}}^{ - \mu }}} \right)\left( {{\mu ^{\text{i}}}} \right)}}{{{\text{i!}}}} $ , given μ = 1.5
 $ \Rightarrow {\text{P}}\left( {{\text{X = i}}} \right) = \dfrac{{\left( {{{\text{e}}^{ - 1.5}}} \right)\left( {{{1.5}^{\text{i}}}} \right)}}{{{\text{i!}}}} $

The probability of no demand, i.e. no car is required is P (X = 0) is given by
 $ \Rightarrow {\text{P}}\left( {{\text{X = 0}}} \right) = \dfrac{{\left( {{{\text{e}}^{ - 1.5}}} \right)\left( {{{1.5}^0}} \right)}}{{{\text{0!}}}} $

Now the probability of some demand reduced, i.e. demand is more than 2 cards which is P (X > 2) is given by
⟹P (X > 2)
⟹1 - P (X ≤ 2)
⟹1 – [P (X = 0) + P (X = 1) + P (X = 2)]
 $
   \Rightarrow 1 - \left[ {\dfrac{{\left( {{{\text{e}}^{ - 1.5}}} \right)\left( {{{1.5}^0}} \right)}}{{{\text{0!}}}} + \dfrac{{\left( {{{\text{e}}^{ - 1.5}}} \right)\left( {{{1.5}^1}} \right)}}{{{\text{1!}}}} + \dfrac{{\left( {{{\text{e}}^{ - 1.5}}} \right)\left( {{{1.5}^2}} \right)}}{{{\text{2!}}}}} \right] \\
   \Rightarrow 1 - {{\text{e}}^{ - 1.5}}\left[ {1 + 1.5 + \dfrac{{2.25}}{2}} \right] \\
   \Rightarrow 1 - {{\text{e}}^{ - 1.5}}\left( {3.625} \right) \\
 $

Therefore the probability that some demand is refused is \[{\text{1 - 3}}{\text{.625 }} \times {\text{ }}{{\text{e}}^{ - 1.5}}\]
Option C is the correct answer.

Note: In order to solve this type of question the key is to figure out the probability that the demand is reduced is nothing but the probability of demand of more than two cars. Also being able to express that the probability of the random variable P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) is a vital step.Formula of poisson distribution is to be known and there is only one parameter for this which is the mean. Figuring this out from the question is also key as it is given indirectly. Poisson distribution is calculated as the probability of an event over an interval. The interval could be anything such as time, area, volume or distance.
Value of the constant e is approximately 2.71828.