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Expert-verified Found in: Page 111 ### A First Course in Probability

Book edition 9th
Author(s) Sheldon M. Ross
Pages 432 pages
ISBN 9780321794772 # Suppose that there are n possible outcomes of a trial, with outcome $i$ resulting with probability $pi,i=1,...,n,\sum _{i=1}^{n}i=1pi=1$. If two independent trials are observed, what is the probability that the result of the second trial is larger than that of the first?

The probability that the first result is larger is equal to the probability that the second result is larger.

The probability that they are equal is $\sum _{i=1}^{n}{p}_{i}^{2}$.

See the step by step solution

## Step 1: Given Information

Trials with results in $\left\{1,2,\dots ,n\right\}$.

$P\left(\text{result is}i\right)={p}_{i}$.

Two independent trials are preformed, with results$\left(i,j\right)$.

## Step 2: Explanation

Firstly, since the trials are independent:

$P\left[\left(i,j\right)\right]={p}_{i}·{p}_{j}$

Therefore

$P\left[\left(i,j\right)\right]=P\left[\left(j,i\right)\right]$

for every $i,j\in \left\{1,2,\dots ,n\right\}$

Now since all $\left(i,j\right)$ for different pairs $i$ and $j$ are mutually exclusive events:

$P\left[\left\{\left(i,j\right);i

And since the result of the two trials can be either that the first result is larger, that the second result is larger or that they are both same (and those possibilities are mutually exclusive)

$1=P\left[\left\{\left(i,j\right);i

## Step 3: Explanation

And that third probability is:

$P\left[\left\{\left(i,i\right),i\in \left\{1,2,\dots n\right\}\right\}\right]=\sum _{i=1}^{n}P\left[\left(i,i\right)\right]=\sum _{i=1}^{n}{p}_{i}^{2}$

Equation (1) states that

$P\left[\left\{\left(i,j\right);i

Therefore the red equation gives:

$1=2P\left[\left\{\left(i,j\right);i

This transforms to:

$P\left[\left\{\left(i,j\right);i

The probability that the first result is larger is equal to the probability that the second result is larger.

Probability that they are equal is $\sum _{i=1}^{n}{p}_{i}^{2}$. ### Want to see more solutions like these? 