## Related questions with answers

You are one of the 10 students performing in a school talent show. The order of the performance is determined at random. The first 5 performers go on stage before the intermission. a. What is the probability that you are the last performer before the intermission and your rival performs immediately before you? b. What is the probability that you are not the first performer?

Solution

VerifiedWe know that the number of permutations of $n$ objects taken $a$ at a time ($a\leq n$), is given as

$\begin{align} P_a^n=\dfrac{n!}{(n-a)!}. \end{align}$

##### a)

We know that in this case the order is important. Also, it's important that you are the last performer before the intermission, so you are fifth performer. On the other hand, it important that your rival performs immediately before you. Therefore, he performs fourth. So, we fix the fourth and fifth place and permutate the remaining eight.

From Equation (1) we get the number of permutations of $n=8$ performers as

$\begin{align*} P_8^8=8! \end{align*}$

This is number of favorable outcomes. On the other hand, the number of possible outcomes is number of permutations of all 10 performers. So, from Equation (1) we get

$\begin{align*} P_{10}^{10}=10! \end{align*}$

Finally, $\textbf{the probability that you are the last performer before the intermission and your rival performs immediately before you}$ is

$\begin{align*} \boxed{P_1=\dfrac{8!}{10!}=\dfrac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\dfrac{1}{10\cdot 9}=\dfrac{1}{90}.} \end{align*}$

##### b)

First, let calculate that you are first performer. As in the part on $a)$ we fix the first place and permutate the remaining $9$ places. That's how we get favorable outcomes and possible outcomes is as in part $a)$, $10!$. So, the probability that you are first performer is

$\begin{align*} P_2^{'}=\dfrac{9!}{10!}=\dfrac{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\dfrac{1}{10}. \end{align*}$

Also we know that for event $A$ and the complement event $\bar{A}$ holds

$\begin{align*} P(A)=1-P(\bar{A}). \end{align*}$

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