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

Short 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-by-step solution

Given Information

Given the values,

probabilities P_$0$, P_$1$ and P_$2$

$\sum _i^2=0$ P_i=$1$

: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,

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

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

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