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Verifying that the probability is equal to 1 Watch

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    A hunter is shooting pheasants and misses with a probability of 75%. He continues hunting until his first hit.
    Find the probability Pn that the hunter hits on the nth shot and verify that the sum(Pn) from n=1 to infinity is 1.

    For the first part I found that Pn = (1/4)^(n-1) * (3/4)^n
    As for the second part I am not sure what I should do with the sum to prove that it converges to 1. I tried rewriting (1/4)^(n-1) * (3/4)^n as (1-3/4)^(n-1) * (3/4)^n and then expanding (1-3/4)^(n-1) part with binomial expansion but it did not seem to help as all of the terms tend to infinity as n tends to infinity.
    I would appreciate any help!
    Thank you in advance!
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    (Original post by spacewalker)
    A hunter is shooting pheasants and misses with a probability of 75%. He continues hunting until his first hit.
    Find the probability Pn that the hunter hits on the nth shot and verify that the sum(Pn) from n=1 to infinity is 1.

    For the first part I found that Pn = (1/4)^(n-1) * (3/4)^n
    As for the second part I am not sure what I should do with the sum to prove that it converges to 1. I tried rewriting (1/4)^(n-1) * (3/4)^n as (1-3/4)^(n-1) * (3/4)^n and then expanding (1-3/4)^(n-1) part with binomial expansion but it did not seem to help as all of the terms tend to infinity as n tends to infinity.
    I would appreciate any help!
    Thank you in advance!
    Check your Pn - it is true for n = 1 but doesn't work for n=2 etc... because for the first hit being on the nth shot, it appears you are multiplying the probability of n index failures by (n-1) indep successes?
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    (Original post by SeanFM)
    Check your Pn - it is true for n = 1 but doesn't work for n=2 etc... because for the first hit being on the nth shot, it appears you are multiplying the probability of n index failures by (n-1) indep successes?
    Oh right, thank you! It should be (3/4)^(n-1)*(1/4). However, I still do not see how the sum of this can give 1 since as n tends to infinity should not (3/4)^(n-1) tend to 1? And so 1 times 1/4 should give 1/4.
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    (Original post by spacewalker)
    Oh right, thank you! It should be (3/4)^(n-1)*(1/4). However, I still do not see how the sum of this can give 1 since as n tends to infinity should not (3/4)^(n-1) tend to 1? And so 1 times 1/4 should give 1/4.
    No, because (and it's helpful to test these things on a calculator eg by finding (3/4)^1000) a value between 0 and 1 raised to a high power tends to 0.

    So, how do you find the sum to infinity of those probabilities?
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    (Original post by SeanFM)
    No, because (and it's helpful to test these things on a calculator eg by finding (3/4)^1000) a value between 0 and 1 raised to a high power tends to 0.

    So, how do you find the sum to infinity of those probabilities?
    Thank you for your reply! (3/4)^(n-1) * (1/4) simplifies to (3/4)^(n-1) * (1-3/4) = (3/4)^(n-1) - (3/4)^n but then from here I am not sure what to do. Could you please give me a hint on what technique I should use?
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    (Original post by spacewalker)
    Thank you for your reply! (3/4)^(n-1) * (1/4) simplifies to (3/4)^(n-1) * (1-3/4) = (3/4)^(n-1) - (3/4)^n but then from here I am not sure what to do. Could you please give me a hint on what technique I should use?
    Gone too far by trying to simplify what you first had, but don't worry, it's all about trying until you work out the right method.

    To give you a hint, what is P(n) when n= 1?

    How about when n=2 and n= 3? Can you see a pattern then see how it makes sense from the general term, and how you can sum it up?
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    (Original post by spacewalker)
    Thank you for your reply! (3/4)^(n-1) * (1/4) simplifies to (3/4)^(n-1) * (1-3/4) = (3/4)^(n-1) - (3/4)^n but then from here I am not sure what to do. Could you please give me a hint on what technique I should use?
    It's a Geometric Progression, for which there's a formula for the sum. Google "Sum of a GP".

    You're actually quite close to being able to work it out yourself.

    If a_n = (3/4)^{n-1} - (3/4)^n, then

    a_1+a_2+...+a_n

    = \underbrace{(3/4)^0 - (3/4)^1}_{a_1} + \underbrace{(3/4)^1 - (3/4)^2}_{a_2}+...+\underbrace{(3/4)^{n-2} - (3/4)^{n-1}}_{a_{n-1}} + \underbrace{(3/4)^{n-1} - (3/4)^{n}}_{a_{n}}

    Look for terms that cancel (hint: nearly everything cancels!) and you're left with

    (3/4)^0 - (3/4)^n (and of course (3/4)^0 = 1).
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    (Original post by SeanFM)
    Gone too far by trying to simplify what you first had, but don't worry, it's all about trying until you work out the right method.

    To give you a hint, what is P(n) when n= 1?

    How about when n=2 and n= 3? Can you see a pattern then see how it makes sense from the general term, and how you can sum it up?
    Thank you! I have just realised to use sum for GP when I went to bed. I have no idea how I could not see that
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    (Original post by spacewalker)
    Thank you! I have just realised to use sum for GP when I went to bed. I have no idea how I could not see that
    Actually I was a bit close minded - see above for how to use method of differences to solve it too.
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    (Original post by SeanFM)
    Actually I was a bit close minded - see above for how to use method of differences to solve it too.
    It's funny, I've never seen the GP sum proved by differences, but I saw what the OP had written in that post (where they'd already made the critical step of splitting each term in the appropriate ratio) and I saw it would work. I think it's quite a cool way of doing it - they should definitely do it as an example in FM when you have a whole topic on method of differences.
 
 
 
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