Book: RD Sharma - Mathematics (Volume 2)
Chapter: 33. Binomial Distribution
Subject: Maths - Class 12th
Q. No. 38 of Exercise 33.1
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In a hospital, there are 20 kidney dialysis machines and that the chance of any one of them to be out of service during a day is 0.02. Determine the probability that exactly 3 machines will be out of service on the same day.
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.
