Q.3.91 - A First Course in Probability

Question

Suppose that n independent trials, each of which results in any of the outcomes $0, 1$, or $2$, with respective probabilities p$0$,p$1$, andp$2$,$\sum _i^2$=0 pi = $1$, are performed. Find the probability that outcomes $1$ and $2$ both occur at least once

Step-by-Step Solution

Answer

The formula of inclusion and exclusion states,

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(AB\right)$

Therefore,

$P\left(E\right)=1-P(A\cup B)$

$=1-\left[P\right(A)+P(B)-P(AB\left)\right]$

$=1-[{(1-p_2)}^n+{(1-p_1)}^n-p_0^n]$

$=1-{(1-p_2)}^n-{(1-p_1)}^n+p_0^n$

Step 1: Given Information

Given the values,

probabilities P_$0$, P_$1$ and P_$2$ 

$\sum _i^{2 }= 0$ P_i =$1$

Step 2 :Explanation

Events:

E = $1$ and $2$ appear at least once in the sequence.

A - each outcome in a sequence is either $0$ or $1$

B - each outcome in a sequence is either  $0$ or $2$

The outcomes of different experiments are independent 

Calculate:P(E)

Start by noting,

$A\cup B=E^c$

Therefore,

$P\left(E\right)=1-P(A\cup B)$

The formula of inclusion and exclusion states,

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(AB\right)$

Step 3: Explanation

The probability that certain experiment will end in either $0$ or $1$, that is $2$, because those events are mutually exclusive is :

$1 - P$_$2$

Because of independence, probability that n experiments end in not $2$ is

$P\left(A\right)={(1-p_2)}^n$

Likewise:

$P\left(B\right)={(1-p_1)}^n$

And AB means that each outcome is $0$ ,because of independence probability of that is:

$P\left(AB\right)=p_0^n$

Therefore:

$P\left(E\right)=1-P(A\cup B)$

$=1-\left[P\right(A)+P(B)-P(AB\left)\right]$

$=1-[{(1-p_2)}^n+{(1-p_1)}^n-p_0^n]$

$=1-{(1-p_2)}^n-{(1-p_1)}^n+p_0^n$

Step 4:Final Answer

The probability that outcomes 1 and 2 both occur at least once is

$=1-{(1-p_2)}^n-{(1-p_1)}^n+p_0^n$