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Math Help - Confusign probability

  1. #1
    Member jacs's Avatar
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    Confusign probability

    A marksman finds that on average he hits the target 9 times out of every 10 and scores a bull's eye on the average once every 5 rounds. He fires 4 rounds. What is the probability that:

    He scores at least 2 bull's eyes and he has hit the target on each of the four rounds?



    I have found that the prob he scores at least 2 bulls eyes to be 0.1808
    and the prob of hittign the target all four times to be 0.6561

    what i am not sure how to do is combine them
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jacs
    A marksman finds that on average he hits the target 9 times out of every 10 and scores a bull's eye on the average once every 5 rounds. He fires 4 rounds. What is the probability that:

    He scores at least 2 bull's eyes and he has hit the target on each of the four rounds?



    I have found that the prob he scores at least 2 bulls eyes to be 0.1808
    and the prob of hittign the target all four times to be 0.6561

    what i am not sure how to do is combine them
    Probability of a b-eye given that the target is hit is 2/9,
    so the probability of 2 b-eyes in 4 hits is:

    <br />
P(2 \mbox{b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 \approx 0.1792<br />

    The probability of four hits from four shots is 0.9^4 \approx 0.6561.

    So the probability of 2 b-eyes and four hits from four shots is:

    <br />
P(2 \mbox{b-eyes and 4 hits from 4 shots})= P(\mbox{2 b-eyes from 4 hits}) P(\mbox{4 hits from 4 shots})
    <br />
\approx 0.1792 \times 0.6561 \approx 0.1176<br />

    RonL
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  3. #3
    Member jacs's Avatar
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    Quote Originally Posted by CaptainBlack
    Probability of a b-eye given that the target is hit is 2/9,
    so the probability of 2 b-eyes in 4 hits is:

    <br />
P(2 \mbox{b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 \approx 0.1792<br />

    The probability of four hits from four shots is 0.9^4 \approx 0.6561.

    So the probability of 2 b-eyes and four hits from four shots is:

    <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br />
P(2 \mbox{b-eyes and 4 hits from 4 shots})= P(\mbox{2 b-eyes from 4 hits}) P(\mbox{4 hits from 4 shots})
    <br /> <br /> <br /> <br /> <br /> <br /> <br />
\approx 0.1792 \times 0.6561 \approx 0.1176<br />





    RonL

    Thanks for that Captain Black, but does it change the probability if it is AT LEAST two bulls-eyes?

    i got the prob for that to be

    <br />
P(2 \mbox{at least 2 b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 <br />

    +{4 \choose 3} (2/9)^3 (7/9)+{4 \choose 4} (2/9)^4\approx 0.1808<br />


    then if i just multiply that with the 0.5661 i get 0.1186 and the text is claiming it is 0.1416

    What i was wondering too, was if you get a bullseye, isnt it implicit that the target is hit? Or am i over thinking this?

    thanks
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    Grand Panjandrum
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    Quote Originally Posted by jacs
    Thanks for that Captain Black, but does it change the probability if it is AT LEAST two bulls-eyes?

    i got the prob for that to be

    <br />
P(2 \mbox{at least 2 b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 <br />

    +{4 \choose 3} (2/9)^3 (7/9)+{4 \choose 4} (2/9)^4\approx 0.1808<br />
    Check you arithmetic, I make this 0.2158

    then if i just multiply that with the 0.5661 i get 0.1186 and the text is claiming it is 0.1416
    You should be multiplying by 0.6561 here.

    What i was wondering too, was if you get a bullseye, isnt it implicit that the target is hit? Or am i over thinking this?
    Yes, but you will need to think about the calculation quite carefully to get
    the right answer.

    RonL
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  5. #5
    Member jacs's Avatar
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    Quote Originally Posted by CaptainBlack
    Check you arithmetic, I make this 0.2158



    You should be multiplying by 0.6561 here.



    Yes, but you will need to think about the calculation quite carefully to get
    the right answer.

    RonL

    yes, my bad, I meant to type 0.6561, got a bit of number dyslexia going there and apparently i cannot press calculator buttons either...grrr....duh to me.... so i had it right all along and just can't calculate properly.

    thanks for that then and sorry for wasting your time.... next time instead of spending ages on the theory, i will double check my button pressing
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  6. #6
    Member jacs's Avatar
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    I think we are looking at the wrong probablity and I just went along for the ride. For at least two two bulls-eyes we must use the bulls-eyes odds which is 1/5, the hitting the target was 2/9.

    i got the prob for that to be

    <br />
P(2 \mbox{at least 2 b-eyes from 4 hits})={4 \choose 2} (1/5)^2 (4/5)^2 <br />

    +{4 \choose 3} (1/5)^3 (4/5)+{4 \choose 4} (1/5)^4\approx 0.1808<br />


    then if i just multiply 0.1808 with the 0.6561 i get 0.1186 and the text is claiming it is 0.1416

    this whole question has just confused me completely and i dont know if i am just misinterpreting the question now

    a) P(hits all four times) = 0.6561 which was correct
    b) P(hits at least 2 bulls-eyes) = 0.1808 which was correct


    c) P(scores at least 2 bulls-eyes and hits all four shots)

    so i figured multiplying them woudl yeild the correct result
    but it doesnt, 0.6561 x 0.1808 = 0.1186

    your method gets that result, 0.6561 x 0.2158 = 0.1416
    but then it would invalidate the previous result in b, wouldnt it?

    or has the text just screwed up???
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by jacs
    I think we are looking at the wrong probablity and I just went along for the ride. For at least two two bulls-eyes we must use the bulls-eyes odds which is 1/5, the hitting the target was 2/9.

    i got the prob for that to be

    <br />
P(2 \mbox{at least 2 b-eyes from 4 hits})={4 \choose 2} (1/5)^2 (4/5)^2 <br />

    +{4 \choose 3} (1/5)^3 (4/5)+{4 \choose 4} (1/5)^4\approx 0.1808<br />


    then if i just multiply 0.1808 with the 0.6561 i get 0.1186 and the text is claiming it is 0.1416
    In order for multiplication to be valid here the two events represented by
    the probabilities have to be independent, but here they are not, because
    the b-eyes imply a hit.

    My approach using the conditional probability of a b-eye gets rid of this
    problem in a systematic manner using:

    <br />
p(a \mbox{ and } b)=p(a \mbox{ given }b).p(b)<br />

    RonL
    Last edited by CaptainBlack; May 22nd 2006 at 08:47 PM.
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  8. #8
    Member jacs's Avatar
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    Quote Originally Posted by CaptainBlack
    In order for multiplication to be valid here the two events represented by
    the probabilities have to be independent, but here they are not, because
    the b-eyes imply a hit.

    My approach using the conditional probability of a b-eye gets rid of this
    problem in a systematic manner using:

    <br />
P(a \mbox{ and } b)=p(a \mbox{ given }b).p(b)<br />

    RonL

    I see, we haven't covered conditional probabilities yet. But i do understand what you mean about the independence, I knew there was some problem there, but i just had no idea what to do about it.

    thank you for all your help
    jacs
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