Question

The mean number of errors due to a bug occurring in a minute is 0.0001. What is the probability that no error will occur in 20 minutes?

Answer 1

This problem is of Poisson Distribution.

We are given, error for one minute is 0.0001

So error for 20 minutes will be = 20*0.0001 = 0.002

Hence $\lambda$ = 0.002

Since Poisson Distribution is given by

$P(X = x) = \frac{ (e^{- \lambda })(\lambda^{x})}{x !}$

Here, x = 0 as there is no error.

Putting values in the equation,

$P(X = x) = \frac{ (e^{- 0.002 })(0.002^{0})}{0 !}$

Solving this we will get

P(x = x) = 0.998001998667 = 0.998