The probability that a particular machine breaks down in any

The probability that a particular machine breaks down in any day is .20 and is independent of he breakdowns on any other day. The machine can break down only once per day. Calculate the probability that the machine breaks down two or more times in ten days. What is the the probability that the machine breaks down for the first time on the third day?

Solution

Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials

probability that a particular machine breaks down in any day is .20

a)
machine breaks down in 10 days rate is 10*0.20 = 2
P( X < 2) = P(X=1) + P(X=0)
= e^-2 * 2 ^ 1 / 1! + e^-2 * ^ 0 / 0!
= 0.406
P( X > = 2 ) = 1 - P (X < 2) = 0.594

b)
probability that the machine breaks down for the first time on the third day = P(no break down for 2 days) + P(breakdown in 3rd day) day

P(no break down for 2 days) = P( X = 0 ), Mean rate be = 2*0.2 = 0.4
P( X = 0 ) = e ^-0.4 * 0.4^0 / 0! = 0.6703

P( Break down in third day) = P(X=1) , Mean rate be 0.20
P( X = 1 ) = e ^-0.2 * 0.2^1 / 1! = 0.1637

P( machine breaks down for the first time on 3rd day) = 0.6703+0.1637 = 0.834