Suppose that n independent trials, each of which results in any of the outcomes 0,1, or 2, with respective probabilities p0,p1, andp2,i2=0 pi = 1, are performed. Find the probability that outcomes 1and2 both occur at least once

Short Answer

Expert verified

The formula of inclusion and exclusion states,

P(AB)=P(A)+P(B)P(AB)

Therefore,

P(E)=1P(AB)

=1[P(A)+P(B)P(AB)]

=1[(1p2)n+(1p1)np0n]

=1(1p2)n(1p1)n+p0n

Step by step solution

01

Given Information

Given the values,

probabilities P0, P1 and P2

i2=0 Pi=1

02

:Explanation

Events:

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

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

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

The outcomes of different experiments are independent

Calculate:P(E)

Start by noting,

AB=Ec

Therefore,

P(E)=1P(AB)

The formula of inclusion and exclusion states,

P(AB)=P(A)+P(B)P(AB)

03

Explanation

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

1-P2

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

P(A)=(1p2)n

Likewise:

P(B)=(1p1)n

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

P(AB)=p0n

Therefore:

P(E)=1P(AB)

=1[P(A)+P(B)P(AB)]

=1[(1p2)n+(1p1)np0n]

=1(1p2)n(1p1)n+p0n

04

Step 4:Final Answer

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

=1(1p2)n(1p1)n+p0n

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