Question

If the number of incoming buses per minute at a bus terminus us a random variable having a Poisson distribution with \[\lambda =0.9\], find the probability that there will be:
(i) exactly \[9\] incoming buses during a period of \[5\] minutes.
(ii) fewer than \[10\] incoming buses during a period to \[8\] minutes.
(iii) at least \[10\] incoming buses during a period of \[11\] minutes.

Answer 1

Hint: In order to find the solution to the given question that is if the number of incoming buses per minute at a bus terminus us a random variable having a Poisson distribution with \[\lambda =0.9\], find the probability that there will be:(i) exactly \[9\] incoming buses during a period of \[5\] minutes. ; (ii) fewer than \[10\] incoming buses during a period to \[8\] minutes. ; (iii) at least \[10\] incoming buses during a period of \[11\] minutes. Apply the formula of Exponential distribution which is \[{{E}_{S}}=\int\limits_{0}^{\infty }{\lambda {{e}^{-\lambda t}}\left[ \left( t+R \right)1\left( t\le s \right)+\left( s+W \right)1\left( t>s \right) \right]}\text{dt}\] where Es​=E (Journey time for strategy ‘\[s\]’) and the journey time is the function of the first arrival time of the rate \[\lambda \] Poisson process of bus arrivals. After this conduct a Poisson experiment, in which the average number of successes within a given region is μ. Then, the Poisson probability is: \[P\left( X;\mu \right)=\dfrac{\left( {{e}^{-\mu }} \right)\left( {{\mu }^{x}} \right)}{x!}\]where \[x\] is the actual number of successes that result from the experiment.

Complete step by step solution:
Let Es​=E (Journey time for strategy ‘\[s\]’). The journey time is the function of the first arrival time of the rate \[\lambda \] Poisson process of bus arrivals. This has Exponential \[\lambda \] distribution. So
\[{{E}_{S}}=\int\limits_{0}^{\infty }{\lambda {{e}^{-\lambda t}}\left[ \left( t+R \right)1\left( t\le s \right)+\left( s+W \right)1\left( t>s \right) \right]}\text{dt}\]
where 1 is the indicator function. Thus
\[{{E}_{S}}=\int\limits_{0}^{s}{\lambda t{{e}^{-\lambda t}}\text{dt}}+R\int\limits_{0}^{s}{\lambda {{e}^{-\lambda t}}\text{dt}}+\left( s+W \right)\int\limits_{0}^{s+W}{\lambda {{e}^{-\lambda t}}\text{dt}}\]
\[\Rightarrow {{E}_{S}}=\dfrac{1-{{e}^{-\lambda s}}}{\lambda }+R\left( 1-{{e}^{-\lambda s}} \right)+W{{e}^{^{-\lambda s}}}\]

Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is μ. Then, the Poisson probability is:

\[P\left( X;\mu \right)=\dfrac{\left( {{e}^{-\mu }} \right)\left( {{\mu }^{x}} \right)}{x!}\]
where \[x\] is the actual number of successes that result from the experiment, and \[e\] is approximately equal to \[2.71828\].


(i) Probability that there will be exactly \[9\] incoming buses during a period of \[5\] minutes means that \[\lambda =4.5\]
\[\Rightarrow P\left( X=9 \right)=\dfrac{{{e}^{-4.5}}\times {{\left( 4.5 \right)}^{9}}}{9!}\]


(ii) Probability that there will be fewer than \[10\] incoming buses during a period to \[8\] minutes means that \[\lambda =7.2\].
Therefore, Required probability \[=\sum\limits_{x=0}^{9}{\dfrac{{{e}^{-7.2}}\times {{\left( 7.2 \right)}^{x}}}{x!}}\]

(iii) Probability that there will be at least \[10\] incoming buses during a period of \[11\] minutes means that \[\lambda =9.9\].
Therefore, Required probability \[=1-\sum\limits_{x=0}^{13}{\dfrac{{{e}^{-9.9}}\times {{\left( 9.9 \right)}^{x}}}{x!}}\]

Note: Students make mistakes while applying the wrong formula for Poisson probability. It’s important to remember that here in this type of question conduct Poisson experiment, in which the average number of successes within a given region is μ. Then, the Poisson probability is: \[P\left( X;\mu \right)=\dfrac{\left( {{e}^{-\mu }} \right)\left( {{\mu }^{x}} \right)}{x!}\]where \[x\] is the actual number of successes that result from the experiment.