# What is the expectation of the number

• Jan 2nd 2010, 04:51 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 $\displaystyle E[X] = \sum x Pr(failure)$

I am stuck here wht should i do
• Jan 2nd 2010, 07:58 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 $\displaystyle p$ of success in each trial?

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

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

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

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

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

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

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

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

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

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

• Jan 3rd 2010, 01:55 AM