Question

There are 5 floors and 3 guards. Each guard can get duty at any of the floors and all 3 can also be posted on the same floor. They are given duty on floors at random. The probability that no two guards are placed on the same floor is…
A. $\dfrac{{12}}{{25}}$
B. $\dfrac{1}{2}$
C. $\dfrac{{20}}{{81}}$
D. $\dfrac{3}{5}$

Answer 1

In such a type of question, probability can be found out by using the concept of the permutation without repetition. Here there are 5 floors and 3 guards and each guard can get duty at any of the floors and all 3 can also be posted on the same floor. So the total number of possible outcomes is $n = 5 \times 5 \times 5 = 125$. Now we have to find the probability that no two guards are placed on the same floor. Possible numbers of combinations are given by ${}^n{P_r}$. As 5 floors and 3 guards, possible number of combinations will be $r = {}^5{P_3}$. Use formula ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$ and $n! = 1 \times 2 \times 3... \times n$ for simplification. As probability is the ratio of possible outcomes to total number of outcomes means $p = \dfrac{r}{n}$, we will obtain the probability that no two guards are placed on the same floor.

Complete step-by-step answer:
Here given that there are 5 floors and 3 guards and each guard can get duty at any of the floors and all 3 can also be posted on the same floor. They are given duty on floors at random.
Here we will first find the total number of outcomes, as each guard can be posted on any of the floors, there are 5 choices for each guard. So the total number of possible outcomes is
$\Rightarrow$ $n = 5 \times 5 \times 5 = 125$.
Now we have to find the probability that no two guards are posted on the same floor means for any 3 of the guards, they can do the duty on any of the 5 floors but not on the same floor.
We will use the concept of permutation without repetition for finding possible outcomes of the event. Possible numbers of combinations are given by ${}^n{P_r}$.
Here there are 5 floors and 3 guards, possible number of combinations will be $r = {}^5{P_3}$.
Now use the formula ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$ for simplification.
$\Rightarrow$ ${}^5{P_3} = \dfrac{{5!}}{{(5 - 3)!}}$
$\Rightarrow$ ${}^5{P_3} = \dfrac{{5!}}{{2!}}$
We know that $n! = 1 \times 2 \times 3... \times n$, so we can write $5! = 1 \times 2 \times 3 \times 4 \times 5$ and $2! = 1 \times 2$.
Putting these values in above form,
${}^5{P_3} = \dfrac{{1 \times 2 \times 3 \times 4 \times 5}}{{1 \times 2}}$
$\Rightarrow$ ${}^5{P_3} = 3 \times 4 \times 5$
$\Rightarrow$ ${}^5{P_3} = 60$.
So, the possible number of outcomes is 60.
As we know that probability is the ratio of possible outcomes to total number of outcomes means $p = \dfrac{r}{n}$.
So, here total outcomes $n = 125$ and possible no of outcomes for an event is $r = 60$.
So probability that no two guards are placed on the same floors is given by $p = \dfrac{r}{n}$.
Putting values, $p = \dfrac{{60}}{{125}}$
Simplifying, $p = \dfrac{{12}}{{25}}$

So, option (A) is the correct answer.

Note: There is an alternative way to solve such a question. As here 5 floors and 3 guards and no two guards can be placed on the same floor so the first guard can be chosen on any of the 5 floors. Second guard can be chosen on balance 4 floors and third guard can be chosen on balance 3 floors as no guards on the same floor. So the total no of possible outcomes is $r = 5 \times 4 \times 3 = 60$. And total no of outcomes is $n = 5 \times 5 \times 5 = 125$. So probability that no two guards are placed on the same floor is $p = \dfrac{r}{n} = \dfrac{{60}}{{125}} = \dfrac{{12}}{{25}}$.