# What is the expectation of the number

• Jan 2nd 2010, 05:51 AM
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
• Jan 2nd 2010, 08:58 AM
Soroban

Quote:

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

• Jan 3rd 2010, 02:55 AM