Answer :

Given that, the chance of any one of the 20 kidney dialysis machines to be out of service during a day is 0.02.


Let p be the probability of a kidney dialysis machine to get out of service.


p = 0.02



Then, q is the probability of a kidney dialysis machine to not be out of service during a day.


And we know that, p + q = 1


q = 1 – p





Let X be a random variable that represents a number of machines out of service during a day out of n machines.


Then, the probability of r machines out of total n machines taken in the sample to get out of service is given by,


P (X = r) = nCrprqn-r


Here, n = 20 (as given in the question)


Putting the values of n, p & q in the above formula, we get


…(i)


We need to find the probability that exactly 3 machines will be out of service on the same day.


So, for this, just put r = 3 in the formula (i),



Observe the formula so obtained after substituting the value of r.


The calculation will be huge since the values of p and q are very small.


So, in this case of low probability events, we use Poisson’s distribution rather than Binomial distribution.


Then,


Poisson’s constant can be found out as,


λ = np


where n = 20 & p = 0.02.


λ = 20 × 0.02


λ = 0.4


Poisson’s distribution is given as,



Put r = 3,



Look up in the table for the value of .




P (X = 3) = 0.0071


, the required probability that exactly 3 machines will be out of service on one day is 0.0071.


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