Question

Each of the six passengers boarded any of the three buses randomly which had no passenger initially. The probability that each bus has got at least one passenger is
$\left( 1 \right)\dfrac{{20}}{{27}}$
$\left( 2 \right)\dfrac{{19}}{{27}}$
$\left( 3 \right)\dfrac{2}{3}$
$\left( 4 \right)\dfrac{5}{9}$

Answer 1

Hint: In this problem, we will consider three cases. Firstly we calculate the total number of ways in which the passengers can board the buses. Next, we will calculate the number of ways in which one of the buses can remain empty. We subtract this value from the total number of ways. On subtracting we subtract twice the number of ways in which two of the buses can remain empty. So we again calculate this value and add it to the expression.

Complete step-by-step answer:
It is given that there are $3$ buses that had no passengers initially and $6$ passengers who want to board any of the three buses.
The total number of ways in which $6$ passengers can board any of the three buses is $ = {3^6}$ .
But over here we have counted those cases where one of the buses remains empty.
But we have a condition that states that each of the buses has at least one passenger.
So we need to subtract those cases from the total number of cases.
The number of ways in which one bus remains empty and two buses has passengers is:
One bus can be chosen in ${}^3{C_1}$ ways and the passengers can board the remaining two buses in ${2^6}$ ways.
The required number of ways is ${}^3{C_1} \times {2^6}$ ways $ = 3 \times {2^6}$ ways.
While subtracting we have subtracted those cases where two of the buses remain empty twice. So we need to add those cases where two buses remain empty.
The number of ways in which two buses remain empty and one bus has all the passengers is:
Two buses can be chosen in ${}^3{C_2}$ ways and the passengers can board the remaining two buses in ${1^6}$ ways.
The required number of ways is ${}^3{C_2} \times {1^6}$ ways $ = 3 \times {1^6}$ ways.
The required number of ways in which the passengers can board the buses that meet the above condition is
$
{3^6} - (3 \times {2^6}) + (3 \times {1^6}) \\
= 729 - (3 \times 64) + 3 \\
= 732 - 192 \\
= 540 \\
$
The required probability of each bus having at least one passenger is
$
\dfrac{{540}}{{{3^6}}} \\
= \dfrac{{540}}{{729}} \\
= \dfrac{{20}}{{27}} \\
$

So, the correct answer is “Option (1)”.

Note: One should first understand the language of the problem. If we can decode the problem in different steps then we can solve it easily. We should be well acquainted with the different methods of calculating the different number of combinations used in the solution.