Question

If 5 distinct balls are placed at random into 5 cells, then the probability that exactly one cell remains empty is
a) $\dfrac{48}{125}$
b) $\dfrac{12}{125}$
c) $\dfrac{8}{125}$
d) $\dfrac{1}{125}$

Answer 1

First we will find the ways of placing 5 distinct balls to place in 5 cells , by using the formula for placing n number of objects in n places, that is ${{n}^{n}}$ ways. Then, we will be selecting one empty cell from 5 cells by using the combination formula that is: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Then, find the number of ways of placing 5 distinct balls into 4 cells such that each cell has at least one ball for that we will first select two balls from five balls then we will assume it as one unit and arrange it in all remaining four boxes by using permutation formula: $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ . Then, finally we will find the probability by using the formula: probability = $\dfrac{\text{number of favourable events}}{\text{number of total events}}$

Complete step by step answer:
We are given in the question that 5 balls are placed into 5 different cells, and we already know that for placing $n$ number of objects in $n$ places, we will have = ${{n}^{n}}$ ways.
Therefore, the number of ways of placing 5 balls in 5 different cells is equal to ${{5}^{5}}$
Now, we have to find the probability that exactly one cell is empty, now we will select one cell out of the given five cells and the selected cell will be empty.
Now we know that selecting $r$ objects from $n$number of objects, we use the formula: ${}^{n}{{C}_{r}}$.
Now, it is given in the question that one cell will be empty therefore, the number of ways of selecting one empty cell out of the five cells is = ${}^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}=5\text{ }......\text{ Equation 1}\text{.}$
Now, if we have to keep exactly one cell empty, then one of the cell must contain 2 balls and the remaining cell must contain one ball each.
So, the number of choosing two balls out of five balls will be: $^{5}{{C}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!}=10\text{ }.......\text{equation 2}\text{.}$
Now let’s assume the two balls as one unit and now in total we have four units that is one from each of the four cells. So, we have to arrange four units in four places therefore we will use permutation: $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}{{\Rightarrow }^{4}}{{P}_{4}}=\dfrac{4!}{\left( 4-4 \right)!}=24\text{ }.........\text{Equation 3}$
So, the total number of ways in which one cell remains empty will be : $^{5}{{C}_{1}}{{\times }^{5}}{{C}_{2}}{{\times }^{4}}{{P}_{4}}$ now putting values from equation 1 , 2 and 3:
We will have the total number of ways in which one cell remains empty is $^{5}{{C}_{1}}{{\times }^{5}}{{C}_{2}}{{\times }^{4}}{{P}_{4}}=5\times 10\times 24=1200\text{ }....\text{Equation 4}$
Now, we know that , probability = $\dfrac{\text{number of favourable events}}{\text{number of total events}}$
So, the required probability is:
$\text{P}=\dfrac{\text{number of ways of placing 5 distinct balls with one cell empty}}{\text{number of ways of placing 5 balls in 5 cells}}$
From equation 4 we know that the total number of ways in which one cell remains empty with five distinct balls is $1200$ , putting this value in the probability formula:
$P=\dfrac{1200}{{{5}^{5}}}=\dfrac{1200}{3125}=\dfrac{48}{125}$

So, the correct answer is “Option A”.

Note: Always remember, while finding the probability of an event you need to find the total number of events and number of favorable events. Hence, the probability is the ratio of favorable events to the total number of events.