Ver good, very good. So when you are at it please solve this problem for me;
You purchase a certain product. The manual states that the lifetime T of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies
P(T≥t)=e−t5, for all t≥0.
For example, the probability that the product lasts more than (or equal to) 2 years is
P(T≥2)=e−25=0.6703.
I purchase the product and use it for two years without any problems. What is the probability that it breaks down in the third year?
Slow down here a minute here, Einstein. You seem to be referring to the Exponential Distribution with probability density function (PDF) P(x) = λe^(−λx) and a cumulative density function (CDF) P(X≤x) = 1 − e^(−λx).
First of all, you need to fix your CDF.
P(T≥2)=e−25=0.6703 should be
For λ=5, P(T≥2)=e^(-2/5)=0.6703.
Secondly, you need to tell theGypsySailor that the PDF is needed to compute the probability of a product breakdown exclusively in the third year. Otherwise, he will be subtracting P(T≥4) from P(T≥3).
With these corrections, I think you are good to go with your problem. Just don't be like that asshole math teacher that the post is complaining about.
IIN that my maths teacher wants our class to fail
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Let me help you out. One plus one is two. Two plus two is four.
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Goldschläger
7 years ago
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flyingnostalgia
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That's very good sailor. Now why did it take you a lifetime to figure this out ?
Ver good, very good. So when you are at it please solve this problem for me;
You purchase a certain product. The manual states that the lifetime T of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies
P(T≥t)=e−t5, for all t≥0.
For example, the probability that the product lasts more than (or equal to) 2 years is
P(T≥2)=e−25=0.6703.
I purchase the product and use it for two years without any problems. What is the probability that it breaks down in the third year?
--
green_boogers
7 years ago
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thegypsysailor
7 years ago
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Slow down here a minute here, Einstein. You seem to be referring to the Exponential Distribution with probability density function (PDF) P(x) = λe^(−λx) and a cumulative density function (CDF) P(X≤x) = 1 − e^(−λx).
First of all, you need to fix your CDF.
P(T≥2)=e−25=0.6703 should be
For λ=5, P(T≥2)=e^(-2/5)=0.6703.
Secondly, you need to tell theGypsySailor that the PDF is needed to compute the probability of a product breakdown exclusively in the third year. Otherwise, he will be subtracting P(T≥4) from P(T≥3).
With these corrections, I think you are good to go with your problem. Just don't be like that asshole math teacher that the post is complaining about.
More info is available at Wikipedia.com
Expectation be damned, the product broke down 3 days after the warranty expired.