20 components and the probability of failure of an individua

clau · **Joined:** Fri, 13 Jul 2012 23:23:44 UTC **Posts:** 2

An electronic system containing 20 components and the probability of failure of an individual component is 0.15.
It is assumed that the components fail intermittently from one another.
a) What is the probability of failure of 2? And at least 1?
b) If one of them knows that has already failed, what is the probability of failure of at least two?
c) If at least one of them has failed, what is the probability of failure of at least two? :confused:

alstat · **Posted:** Sat, 14 Jul 2012 02:49:57 UTC

I'll get you started on the first one:

Number of ways to choose 2 bad components out of 20 = ${20 \choose 2}$ .

That leaves 18 good ones (P=.85)

, so

$P(2) = {20 \choose 2}(.15^2)(.85^{18}) \approx .229$ .

Now for $P(\geq 1)$ , you could add P(1) + P(2) + ... + P(20)

, but there's an easier way by first finding P(0)...

outermeasure · **Posted:** Sat, 14 Jul 2012 02:50:23 UTC

clau wrote:

An electronic system containing 20 components and the probability of failure of an individual component is 0.15.
It is assumed that the components fail intermittently from one another.
a) What is the probability of failure of 2? And at least 1?
b) If one of them knows that has already failed, what is the probability of failure of at least two?
c) If at least one of them has failed, what is the probability of failure of at least two? :confused:

What do you mean by "components fail intermittently from one another"?

clau · **Joined:** Fri, 13 Jul 2012 23:23:44 UTC **Posts:** 2

Forgiveness is not intermittent is independent, the most important question that I need are b) If one of them knows that has already failed, what is the probability of failure of at least two?
c) If at least one of them has failed, what is the probability of failure of at least two?

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20 components and the probability of failure of an individua

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