Q7.28 - A First Course in Probability

Question

Suppose that a sequence of independent trials in which each trial is a success with probability p is performed until either a success occurs or a total of n trials has been reached. Find the mean number of trials that are performed. Hint: The computations are simplified if you use the identity that for a nonnegative integer-valued random variable X, 

$E\left[X\right]=\sum _{i=1}^{\mathrm{\infty }} P\{X\ge i\}$

Step-by-Step Solution

Answer

The mean number of trials that are performed $\frac{1-q^n}{p}$.

Step 1: Given information

A sequence of independent trials in which each trial is a success with probability p is performed.

Step 2: Solution

The probability mass function of X is 

$P(X=i)=q^{i-1}p\mathrm{\forall }i=1,2,\dots$

Now lets consider the $P(X\ge i)=\sum _{x=i}^{\mathrm{\infty }} P(X=x)$

$=\sum _{x=i}^{\mathrm{\infty }} q^{x-1}p$

$=p\left[q^{i-1}+q^i+q^{i+1}+\dots \right]$

$=p\left[q^{i-1}\left(1+q+q^2+\dots \right)\right]$

$=p{q}^{i-1}(1-q{)}^{-1}$

Therefore, $1+q+q^2+\dots$ is an infinite geometric series

$=q^{i-1}$

Step 3: Solution

Now we need to calculate 

$\mathrm{E}\left[\mathrm{X}\right]=\sum _{i=1}^{\mathrm{\infty }} P(X\ge i)$

$=\sum _{i=1}^n P(X\ge i)$

number of trials is finite (n)

$=\sum _{i=1}^n q^{i-1}$

$=1+q+q^2+\dots +q^{n-1}$

$=\frac{1-q^n}{1-q}$

$1+q+q^2+\dots +q^{n-1}$ is an finite geometric series

$=\frac{1-q^n}{p}$

Step 4: Final answer

The mean number of trials that are performed $\frac{1-q^n}{p}$̣