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
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.
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.
Rate this question :
A box has 100 penMathematics - Exemplar
State True or FalMathematics - Exemplar
A random variableMathematics - Board Papers
How many times muMathematics - Board Papers
A card from a pacMathematics - Board Papers
An experiment sucMathematics - Board Papers
A bag A containsMathematics - Board Papers
On a multiple choMathematics - Board Papers
Mark the correctRD Sharma - Volume 2