Question

A car hire firm has two cars which it hires out day by day. The number of demands for a car on each day is distributed as Poisson variate X with mean 1.5. The probability that some

demand is refused is –

• A. 9/8e^1**.**5
• B. 9/8e
• C. 9/8e²
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

We have X ~ P (l) where l = 1.5. Thus

P (Some demand is refused) = P (X > 2)

= 1 – P (X £ 2) = 1 – e^–l $\left[ 1 + \lambda + \frac { \lambda ^ { 2 } } { 2 ! } \right]$

= 1 – e^–1**.**5 $\left[ 1 + 1.5 + \frac { ( 1.5 ) ^ { 2 } } { 2 } \right]$

= 1 – e^–1**.**5 (3.625)