If the length of time the computer part lasts is exponentially distributed with mean value is $10$.

So, for the exponential distribution, we can find the probability of one computer parts.

$$p(x>7) = e^{(-m * 7)} = 0.4966$$ where $m = \frac{1}{mean} = 0.1$.

My question, what is probability of $100000$ computer parts lasts more than seven years ?

Assuming their lifespan is independepent, and they are all exponentially distributed like you say,

$P(\text{all parts make it}) = \prod_{p\in\text{all products}}\exp(-m\cdot7) = \exp(-m\cdot 7)^{100000}$

First step is independence, second step is rewriting the same thing.

It's just the number you have, 0.49, multiplied with itself 100000 times, $$0.49\cdot0.49\cdot\ \dots\ \cdot0.49$$.

July 12, 2019 08:44 AM

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