Question

Let,X₁, X₂andX₃and be independent lifetimes of memory chips. Suppose that each\({X_i}\)has the normal distribution with mean 300 hours and standard deviation 10 hours. Compute the probability that at least one of the three chips lasts at least 290 hours.

Short answer

Probability that at least one of the three chips lasts at least 290 hours is 0.996

Step-by-step solution

Given information

X₁, X₂andX₃ are independent lifetimes of memory chips.

X_i has the normal distribution with mean 300 hours and standard deviation 10 hours.

Computing the probability

\({\rm{Ai}}\)be the event that chip i lasts at most 290 hours.

We want the probability of\(\bigcup\limits_{i = 1}^3 {A_i^c} \), whose probability is:

\(1 - {\rm P}\left( {\bigcap\limits_{i = 1}^3 {{A_i}} } \right) = 1 - \prod\limits_{i = 1}^3 {{\rm P}\left( {{A_i}} \right)} \)

Since the lifetime of each chip has the normal distribution with 300 and standard deviation 10,each\({A_i}\)has probability.

\(\begin{array}{c}\phi \left( {\frac{{\left[ {290 - 300} \right]}}{{10}}} \right) = \phi \left( { - 1} \right)\\ = 1 - 0.8413\\ = 0.1587\end{array}\)

So the required probability is -

\(\begin{array}{l}1 - {0.1587^3}\\ = 0.9960\end{array}\)

Hence, the probability that at least one of the three chip lasts at least 290 hours is 0.996