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Math Help - Lottery Ticket Word Problem

  1. #1
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    Lottery Ticket Word Problem

    Hey I'm kind of stuck with this:

    1 out of 10 scratch off lottery tickets wins.

    If you buy 15 tickets what is the probability that at least three win a prize?
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  2. #2
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    Quote Originally Posted by Flipz4226 View Post
    Hey I'm kind of stuck with this:

    1 out of 10 scratch off lottery tickets wins.

    If you buy 15 tickets what is the probability that at least three win a prize?
    This is a binomial probability distribution question with p=.1 and n=15

    Can you solve it now?

    Your calc might have a function, or you can type in the summation, if doing it by hand you should not that P(at least 3 win)=1-P(0 win or 1 wins or 2 win)
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  3. #3
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    Right, so I got: 1 - (.206+.023+.267) = .504

    What about if I want to know the least amount of tickets to buy so the probability of winning at least one ticket is greater than .8? Thanks
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  4. #4
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    your P(1 win) is incorrect which is throwing off your answer

    use the binomial again, but you'll be solving an equation for n, the number of tickets to buy, you want the sum to be >= .8
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  5. #5
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    I thought f(1) looked wrong so: 1 - (.206+.343+.267) = .184

    I have: .8 >= n Sumation k=0 (n | k) * (.1^k) * (.9^n-k) is this how I should solve for n? Thanks a lot.
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  6. #6
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    Quote Originally Posted by Flipz4226 View Post
    I thought f(1) looked wrong so: 1 - (.206+.343+.267) = .184

    I have: .8 >= n Sumation k=0 (n | k) * (.1^k) * (.9^n-k) is this how I should solve for n? Thanks a lot.
    I'm not exactly sure what you mean by (n|k) i think you are correct, but let me make it clearer

    \sum_{k=1}^x {{x}\choose{k}}.1^k\cdot .9^{x-k}\geq .8

    on second thought, this is a complicated equation to solve, and nothing too enlightening or complex is going on here.... if you are just using a calculator i would just keep increasing n by 1 until you get the probability to be >=.8

    if you have a graphing calculator you should have a binomialcdf command that you can take advantage of... I am getting 16 as the answer
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  7. #7
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    What are you using as the parameters? I'm doing 1-binomcdf(29,.1,0) = .952 for P(X>=1). So I'm getting 29 tickets. But I feel like something is being done wrong because this is supposed to be solved by hand.
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  8. #8
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    Quote Originally Posted by Flipz4226 View Post
    What are you using as the parameters? I'm doing 1-binomcdf(29,.1,0) = .952 for P(X>=1). So I'm getting 29 tickets. But I feel like something is being done wrong because this is supposed to be solved by hand.
    Do 1-binompdf(16,.1,0) and then do 1-binompdf(15,.1,0) you'll see that you need 16

    ooor you could do

    binomcdf(15,.1,1,15)=.79 and then binomcdf(16,.1,1,16)>.8

    solving by hand seems pointless and unreasonable, but if you insist on doing it by hand I'm not sure I have any tips for you
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