Question

(A9). 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

Let X = the random variable denoting the number of errors due to the bug occurring in a minute

The values of X being 0, 1, 2, ..... $\infty$

The mean number of errors due to the bug occurring in a minute = 0.0001

Then, X ~ Poisson($\lambda$ = 0.0001) [since, for Poisson distribution parameter = mean]

Then the p.m.f. of X is given by,

P(X = x) = $\begin{cases} \frac{e^{-0.0001}.(0.0001)^x}{x!} &\text{for } x=0,1,2,...\infty \\ 0 &\text{otherwise } \end{cases}$

Now, the probability that no error will occur in 1 minute

= P(X = 0)

= $\frac{e^{-0.0001}.(0.0001)^0}{0!}$

= e^-0.0001

$\therefore$ The probability that no error will occur in 20 minutes

= $\displaystyle\prod_{i=1}^{20}$ (The probability that no error will occur in i^th minute)

= [P(X = 0)]²⁰

= [e^-0.0001]²⁰

= e^-0.002

= 0.998 (rounded to 3 decimal places)

Answer: The probability that no error will occur in 20 minutes is 0.998.