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Math Help - Conditional lifetimes

  1. #1
    Lord of certain Rings
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    Conditional lifetimes

    Hi everyone,

    I am slowly learning the basics of probability theory. I have realized that learning theorems on continuity of probability and rigorous proofs is one thing, but solving problems is something totally different

    Problem:
    Let X represent the lifetime, rounded up to an integer number of years, of a certain car battery.Suppose that the pmf of X is given by p_X (k) = 0.2 if 3 \leq k \leq 7 and p_X (k) = 0 otherwise.

    (i)Find the probability, P [X > 3], that a three year old battery is still working.

    (ii) Given that the battery is still working after five years, what is the conditional probability that the battery will still be working three years later? (i.e. what is P [X > 8|X > 5]?)

    My idea:

    (i)I thought this was straightforward and noted that it is a discrete distribution. So

    P[X > 3] = \sum_{k=4}^{k=\infty} p_X(k) = 1 - \sum_{k=-\infty}^{k=3} p_X(k) = 1 - (0.2) = 0.8

    (ii) Here I am stumped. The lifetime of the battery has an equally likely chance of failing from 3rd year to 7th year. How can the battery still work at around the 8th year?!

    I have most likely misunderstood the problem. Moreover this problem is to motivate "geometric" distribution!

    Any help appreciated

    Thanks,
    Srikanth
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  2. #2
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    Quote Originally Posted by Isomorphism View Post
    Hi everyone,

    I am slowly learning the basics of probability theory. I have realized that learning theorems on continuity of probability and rigorous proofs is one thing, but solving problems is something totally different

    Problem:
    Let X represent the lifetime, rounded up to an integer number of years, of a certain car battery.Suppose that the pmf of X is given by p_X (k) = 0.2 if 3 \leq k \leq 7 and p_X (k) = 0 otherwise.

    (i)Find the probability, P [X > 3], that a three year old battery is still working.

    (ii) Given that the battery is still working after five years, what is the conditional probability that the battery will still be working three years later? (i.e. what is P [X > 8|X > 5]?)

    My idea:

    (i)I thought this was straightforward and noted that it is a discrete distribution. So

    P[X > 3] = \sum_{k=4}^{k=\infty} p_X(k) = 1 - \sum_{k=-\infty}^{k=3} p_X(k) = 1 - (0.2) = 0.8 Mr F says: This looks good to me.

    (ii) Here I am stumped. The lifetime of the battery has an equally likely chance of failing from 3rd year to 7th year. How can the battery still work at around the 8th year?!

    I have most likely misunderstood the problem. Moreover this problem is to motivate "geometric" distribution!

    Any help appreciated

    Thanks,
    Srikanth
    (ii) is an odd one in light of your final remark .... \Pr(X \geq 8 | X \geq 5) = \frac{\Pr(X \geq 8 ~ \text{and} ~ X \geq 5)}{\Pr(X \geq 5)} = \frac{\Pr(X \geq 8)}{\Pr(X \geq 5)} = 0 I would have thought ......
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  3. #3
    Lord of certain Rings
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    But Mr.F, this is supposed to be a geometric distribution problem. \Pr(X \geq 8 | X \geq 5) was supposed to be \Pr(X \geq 3) and thus we would have seen the memoryless property of the geometric distribution.

    I cannot understand the problem, yet there are two pieces of observations I can make:

    (*)The idea of a lifetime can be related to success and failure somehow, so we could use it as a repeated Bernoulli trial.

    (**) The number of times we have to repeat a Bernoulli experiment until we get our first success is a geometric distribution.So the battery surviving after 8 years is like no success(here success is battery failing).

    Can I use this somehow, given that you could make out more from the problem?

    Thanks,
    Srikanth
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  4. #4
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    Quote Originally Posted by Isomorphism View Post
    But Mr.F, this is supposed to be a geometric distribution problem. \Pr(X \geq 8 | X \geq 5) was supposed to be \Pr(X \geq 3) and thus we would have seen the memoryless property of the geometric distribution.

    I cannot understand the problem, yet there are two pieces of observations I can make:

    (*)The idea of a lifetime can be related to success and failure somehow, so we could use it as a repeated Bernoulli trial.

    (**) The number of times we have to repeat a Bernoulli experiment until we get our first success is a geometric distribution.So the battery surviving after 8 years is like no success(here success is battery failing).

    Can I use this somehow, given that you could make out more from the problem?

    Thanks,
    Srikanth
    The given definition is:
    Let X represent the lifetime, rounded up to an integer number of years, of a certain car battery.Suppose that the pmf of X is given by if and otherwise.
    What is k meant to be ......? It's obviously NOT the possible values of X (which is what I assumed in my previous reply) .....

    Is k meant to represent the number of years that the battery will last after any given number of years ....? If so, then clearly \Pr(X \geq 8 | X \geq 5) = \Pr(X \geq 3) .....
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  5. #5
    Lord of certain Rings
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    Quote Originally Posted by mr fantastic View Post
    The given definition is:

    What is k meant to be ......? It's obviously NOT the possible values of X (which is what I assumed in my previous reply) .....

    Is k meant to represent the number of years that the battery will last after any given number of years ....? If so, then clearly \Pr(X \geq 8 | X \geq 5) = \Pr(X \geq 3) .....
    Yes you are right. The title was misleading and thats why I was worried. Thanks
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