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Thread: Binomial Problem

  1. #1
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    Binomial Problem

    Ok yes the Actex Manual has the solution, but I don't understand the solution so i would appreciate some assistance. Here is the question:

    A carnival Sharpshooter game charges $25 for 25 shots at a target. If the shooter hits the bullseye fewer than 5 times then he gets no prize. If he hits the bullseye 5 times he gets back $10. For each additional bullseye over 5 he gets back an additional $5. The shooter estimates that he has a .2 probability of hitting the bullseye on any given shot. What is the shooter's expected gain if he plays the game (nearest $1)?

    Ok they set up the Expected prize as

    E(5x-15) + 15.p(0) + 10.p(1) + 5.p(2) + 0.p(3) - 5.p(4)

    why is 5.p(4) subtracted???

  2. #2
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    I haven't looked at it too long but these problems/solutions tend to be given in "layers" so you subtract to account for the layer where you get 0. Also, did you mean to have an "=" somewhere in there?

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    Quote Originally Posted by hershey79 View Post
    Ok yes the Actex Manual has the solution, but I don't understand the solution so i would appreciate some assistance. Here is the question:

    A carnival Sharpshooter game charges $25 for 25 shots at a target. If the shooter hits the bullseye fewer than 5 times then he gets no prize. If he hits the bullseye 5 times he gets back $10. For each additional bullseye over 5 he gets back an additional $5. The shooter estimates that he has a .2 probability of hitting the bullseye on any given shot. What is the shooter's expected gain if he plays the game (nearest $1)?

    Ok they set up the Expected prize as

    E(5x-15) + 15.p(0) + 10.p(1) + 5.p(2) + 0.p(3) - 5.p(4)

    why is 5.p(4) subtracted???
    I'm not sure how you got that. From what I recall,

    Let x = number of bullseye

    E(gains) = (-25)p(x=0,x=1,x=2,x=3,x=4) + (-15)p(x=5) + (-10)p(x=6) + ... + (80)p(x=25)

    Solving for each p(x) and you get your expected gain

  4. #4
    Actuary.com - Level III Poster mathrix's Avatar
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    Quote Originally Posted by hershey79 View Post
    Ok yes the Actex Manual has the solution, but I don't understand the solution so i would appreciate some assistance. Here is the question:

    A carnival Sharpshooter game charges $25 for 25 shots at a target. If the shooter hits the bullseye fewer than 5 times then he gets no prize. If he hits the bullseye 5 times he gets back $10. For each additional bullseye over 5 he gets back an additional $5. The shooter estimates that he has a .2 probability of hitting the bullseye on any given shot. What is the shooter's expected gain if he plays the game (nearest $1)?

    Ok they set up the Expected prize as

    E(5x-15) + 15.p(0) + 10.p(1) + 5.p(2) + 0.p(3) - 5.p(4)

    why is 5.p(4) subtracted???

    X - # of bullseye ~ Binomial ( n=25, p=.2)

    X = 0 1 2 3 4 5 6 7 . . . 25
    5X-15 = -15 -10 -5 0 5 10 15 20 . . . 110

    The question asks for E[5X-15] but not including those X=0,1,2,3,4

    Bottomline: Look at the signs of 5X-15 on those X's

    You should be able to figure out. :smiloe:

  5. #5
    Actuary.com - Level III Poster mathrix's Avatar
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    Quote Originally Posted by KHC831 View Post
    I'm not sure how you got that. From what I recall,

    Let x = number of bullseye

    E(gains) = (-25)p(x=0,x=1,x=2,x=3,x=4) + (-15)p(x=5) + (-10)p(x=6) + ... + (80)p(x=25)

    Solving for each p(x) and you get your expected gain
    Possible but good luck on doing this

  6. #6
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    ''' thanks for the inputs!

    I will take again a look at those signs.

    Hershey79

  7. #7
    Actuary.com - Level III Poster mathrix's Avatar
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    Quote Originally Posted by hershey79 View Post
    ''' thanks for the inputs!

    I will take again a look at those signs.

    Hershey79
    Note: E[X-15] is the expected prize not yet the answer on the question which asks for the expected gain.

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