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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 $m = \frac{1}{mean} = 0.1$.

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

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  • $\begingroup$ The probability all $100000$ parts last $7$ years? Very very small. There is only a $0.1\%$ chance they will all last $6$ hours $\endgroup$ – Henry Jul 12 at 10:01
  • $\begingroup$ @Henry, thanks, could you please explain with solution? $\endgroup$ – dtc348 Jul 12 at 10:31
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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, $$0.49\cdot0.49\cdot\ \dots\ \cdot0.49$$.

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  • $\begingroup$ Thanks, if we include total lifespan is 100000, then, how would be the result looks like? $\endgroup$ – dtc348 Jul 12 at 8:58
  • $\begingroup$ Result is 0, if $0.49^{100000}$ $\endgroup$ – dtc348 Jul 12 at 10:36
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    $\begingroup$ Yes, it's zero for a computer (due to rounding) and for all practical purposes. $\endgroup$ – Gijs Jul 12 at 11:10

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