Question

Each of the $n$ passengers sitting in a bus may get down from it at the next stop with probability $p$. Moreover, at the next stop either no passenger or exactly one passenger boards the bus. The probability of no passenger boarding the bus at the next stop being ${{p}_{o}}$. Find the probability that when the bus continues on its way after the stop, there will again be $n$ passengers in the bus.
A. ${{\left( 1-p \right)}^{n-1}}.\left[ {{p}_{o}}\left( 1-p \right)+np\left( 1-{{p}_{o}} \right) \right]$
B. $1-{{\left( 1-p \right)}^{n-1}}.\left[ {{p}_{o}}\left( 1-p \right)+np\left( 1-{{p}_{o}} \right) \right]$
C. ${{p}^{n}}.\left[ {{p}_{o}}\left( 1-p \right)+np\left( 1-{{p}_{o}} \right) \right]$
D. $1-{{p}^{n}}.\left[ {{p}_{o}}\left( 1-p \right)+np\left( 1-{{p}_{o}} \right) \right]$

Answer 1

Hint: In this problem we will take two cases and find the probability of each case and add them to get the required result. We will take the first case as ‘No passenger gets down at next stop and no passenger boards the bus.’ In this case there is no change in the number of passengers in the bus. So, we can calculate the probability for this case using the probability distribution factor. For the second case we will take ‘only one passenger gets down at next stop and one passenger boards the bus.’ In this case also there is no change in the number of passengers in the bus. Here also we will calculate the probability of this case using the probability distribution formula and add the both the probabilities to get the required probability.

Complete step-by-step answer:
Given that,
Number of passengers in bus is $n$.
Probability of getting down passenger in next stop is $p$.
The event not getting down passenger in next stop is the complementary event for the event that getting down passenger in next stop. So, the probability of not getting down passenger in next stop is $1-p$.
Probability of no passenger boarding in the next stop is ${{p}_{o}}$.
Now the probability of boarding one passenger in the next stop is complementary event for the event of no passenger boarding in the next stop. So, the probability of boarding one passenger in the next stop is $1-{{p}_{o}}$.

Case-1:
Now the probability of no passenger gets down at next stop and no passenger boards the bus is given by
$P\left( {{n}_{1}} \right)={{\left( 1-p \right)}^{n}}{{p}_{o}}$.
Case-2:
Now the probability of only one passenger gets down at next stop and one passenger boards the bus is given by
$P\left( {{n}_{2}} \right)={}^{n}{{C}_{1}}p{{\left( 1-p \right)}^{n-1}}\left( 1-{{p}_{o}} \right)$
Now the required probability is given by
$\begin{align}
  & P\left( n \right)=P\left( {{n}_{1}} \right)+P\left( {{n}_{2}} \right) \\
 & \Rightarrow P\left( n \right)={{\left( 1-p \right)}^{n}}{{p}_{o}}+{}^{n}{{C}_{1}}{{\left( 1-p \right)}^{n-1}}\left( 1-{{p}_{o}} \right) \\
 & \Rightarrow P\left( n \right)={{\left( 1-p \right)}^{n-1}}\left[ {{p}_{o}}\left( 1-p \right)+np\left( 1-{{p}_{o}} \right) \right] \\
\end{align}$

So, the correct answer is “Option A”.

Note: The problem comes with a hard language one should clearly read the problem and should be able to list out all the parameters given and what we need to calculate. Once you understand the problem’s language you can simply solve the problem.