Question

Let\({X_1},{X_2},{X_3}\)be a random sample from the exponential distribution with parameter\(\beta \). Find the probability that at least one of the random variables is greater than t, where\(t > 0\)

Short answer

Probability that at least one of the random variables is than t is\[1 - {\left[ {1 - \exp \left( { - \beta t} \right)} \right]^3}\].

Step-by-step solution

Given information

\({X_1},{X_2},{X_3}\) be the random samples from exponential distribution with parameter \(\beta \).

Computing the probability

Let A_i = {X_i > t} for i = 1,2,3

The event that at least one X_i is greater than t is \[\bigcup\limits_{i = 1}^3 {{A_i}} \]

By using\[{\rm P}\left( {\bigcup\limits_{i = 1}^3 {{A_i}} } \right) = 1 - {\rm P}\left( {\bigcap\limits_{i = 1}^3 {A_i^c} } \right)\].

The X_i are mutually independent and identically distributed.

\[\begin{array}{c}{\rm P}\left( {\bigcap\limits_{i = 1}^3 {A_i^c} } \right) = {\rm P}{\left( {A_1^c} \right)^3}\\ = {\left[ {1 - \exp \left( { - \beta t} \right)} \right]^3}\end{array}\]

So probability we want to calculate is\[1 - {\left[ {1 - \exp \left( { - \beta t} \right)} \right]^3}\].

Hence probability that at least one of the random variable is t is \[1 - {\left[ {1 - \exp \left( { - \beta t} \right)} \right]^3}\].