Question

Part 1

The probability of observing a certain event is 0.2. Suppose we make repeated observations until we have observed the desired event three times. What is the probability that we make five observations in order to have observed the desired event three times.? (Round your answer to four decimal places.)

Part 2

A probability of failure on any given trial is given as 0.01.

Use the Poisson approximation to the binomial to find the (approximate) probability of at least five failures in 200 trials? (Round your answer to three decimal places.)

Answer 1

Part 1. Given the probability of observing a certain event is p = 0.2,

Then probability of not observing the certain event is q=1-p = 1-0.2 = 0.8

we make repeated observations until we have observed the desired event three times

Then the probability that we make five observations in order to have observed the desired event three times is found as follows:

Probability that we make five observations in order to have observed the desired event three times =

=P(In 4 observations the desired event occurs 2 times and in the last 5th observation the desired event occurs the last third time)

= P(In 4 observations the desired event occurs 2 times)*P(in the last 5th observation the desired event occurs the last third time)

= P(In 4 observations the desired event occurs 2 times in any 4C2 ways and the remaining 2 times the desired event does not occur)*p      

(since, probability of observing a desired event one time = p)

=4C2*p^2*q^2*p

=4C2*p^3*q^2

=6*(0.2)^3*(0.8)^2

=6*0.008*0.64

=0.0307

Note: The above probability can also be found by using negative binomial distribution probability mass function as

P(X=x)=x+r-1Cr-1(p^r)(q^x) for x=2 and r=3, then

P(X=2) =2+3-1C3-1*p^3*q^2 = 4C2*p^3*q^2 = 6*(0.2)^3*(0.8)^2=0.0307.

Part 2 The probability of failure on any given trial is given as p= 0.01

n= number of trials = 200

Then the random variable X as number of failures follows Poisson distribution with parameter m=np = 0.01x200 = 2

P(X=r) =[e^(-m)]x[m^r]/r! =[e^(-2)]x[2^r]/r!, r=0,1,2,3,….

Then the probability of at least five failures in 200 trials is

P(X≥5) = 1- P(X<5) = 1 – P(X≤4)

= 1- e^(-2)[1+[2^1]/1!+ [2^2]/2!+ [2^3]/3! +[2^4]/4!]

=1-0.135[1+(2/1)+(4/2)+(8/6)+(16/24)]

=1-0.135[1+2+2+1.333+0.666]

=1-0.135[6.999]

=1-0.944

=0.056.