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) = ⁿC_rp^rq^n-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.