# Math Help - What is the expectation of the number

1. ## What is the expectation of the number

What is the expectation of the number of failure preceding first success in an infinite series of independent trials with constant probability p of success in each trial?

Solution

Pr(success) = p
Pr(failure ) = 1-p

Expectation of failure $E[X] = \sum x Pr(failure)$

I am stuck here wht should i do

What is the expectation of the number of failures preceding first success
in an infinite series of independent trials with constant probability $p$ of success in each trial?

Let: . $S = \text{success},\;F = \text{failure}$

We have: . $\begin{array}{ccc}P(S) &=& p \\ P(F) &=& q \end{array} \qquad \text{where }p+q\,=\,1$

$\begin{array}{ccc}
\text{no. of F} & \text{scenario} & \text{Probabiity} \\ \hline
0 & S & p \\
1 & FS & qp \\
2 & FFS & q^2p \\
3 & FFFS & q^3p \\
4 & FFFFS & q^4p \\
\vdots & \vdots & \vdots
\end{array}$

Hence: . $E \;=\;0(p) + 1(qp) + 2(q^2p) + 3(q^3p) + 4(q^4p) + \hdots$

$\begin{array}{cccc}
\text{We have:} & E & = & qp + 2q^2p + 3q^3p + 4q^4p + 5q^5p + \hdots \\
\text{Multiply by }q: & qE &=& \qquad q^2p + 2q^3p + 3q^4p + 4q^5p + \hdots \end{array}$

. . $\text{Subtract: }E - qE \;=\; qp + q^2p + q^3p + q^4p + q^5p + \hdots$

$\text{We have: }\;\;E(1-q) \;=\;qp\underbrace{(1 + q + q^2 + q^3 + \hdots)}_{\text{geometric series}}$
. . The sum of the geometric series is: . $\frac{1}{1-q}$

So we have: . $E(1-q) \:=\:qp\cdot\frac{1}{1-q}$
. . Then: . $E \;=\;\frac{pq}{(1-q)^2}$

But $q \,=\,1-p$

. . Therefore: . $E \;=\;\frac{p(1-p)}{p^2} \;=\;\boxed{\frac{1-p}{p}}$

3. why did u subtract the 2 equations

why did u subtract the 2 equations
It is a standard trick used to find the sum of a geometric series. You are strongly advised to review geometric series - the question assumes you understand that material.

In fact, you are strongly advised to review all of the material that you are meant to have learned in your previous years of study.