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anonymous

  • 5 years ago

A pair of fair dice is tossed 3 times. Find the probability a seven appears 3 times.

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  1. anonymous
    • 5 years ago
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    i'm gonna go with 16.67%^3=0.00463241...

  2. anonymous
    • 5 years ago
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    probability that seven appears on one throw = 6/36 = 1/6 probability that it appears 3 times = 1/6 *1/6*1/6 = 1/216

  3. anonymous
    • 5 years ago
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    So, as a percentage, 0.463%

  4. anonymous
    • 5 years ago
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    thats correct but what happens when if they ask you to Find the probability a 7 or 11 appears at least twice.

  5. anonymous
    • 5 years ago
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    it is a binomial probability... probability that 7 or 11 appear on one throw = 8/36 = 2/9 probability that 7or 11 appear at least 2 times = 3C2 * (2/9)^2 * (7/9)

  6. anonymous
    • 5 years ago
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    P(7) =0 unless you're looking at the probability that the sum is 7...but a fair die has numbers 1 to 6 only...no 7...thus P(7) = 0 for each toss...and thus 0 overall...

  7. anonymous
    • 5 years ago
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    GODinme it is the sum of 7 or 11

  8. anonymous
    • 5 years ago
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    sorry, it is probability that 7 or 11 appear exactly 2 times. for at least 2 times please add this also... 3C3 *(2/9)^3

  9. anonymous
    • 5 years ago
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    ok...that wasn't in the question...who's asking the question?

  10. anonymous
    • 5 years ago
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    7 or 11? Thats a Craps problem...Google will help...

  11. anonymous
    • 5 years ago
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    heartslayer20 asked it..........

  12. anonymous
    • 5 years ago
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    so how do you know what it's sposed to be?

  13. anonymous
    • 5 years ago
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    it was a two part question you know it is a sum because of course there isnt a 7 or a 11. The question was a two parts. I got the first answer from the first two replies. the second part is "A pair of fair dice is tossed 3 times.Find the probability a 7 or 11 appears at least twice.

  14. anonymous
    • 5 years ago
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    probability that 7or 11 appear at at least 2 times = 3C2 * (2/9)^2 * (7/9) + 3C3 *(2/9)^3

  15. anonymous
    • 5 years ago
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    probability of a 7, 1/6th probability of an 11, 1/18th. Multiply ad nauseum...

  16. anonymous
    • 5 years ago
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    ok...then it can be represented as a binomial distribution: n=3, P(7) = 1/6 for any throw...and each throw is independent... thus: 3C3*(1/6)^3*(5/6)^0 + 3C2*(1/6)^2*(5/6)^1 P(11) = 1/18 so similarly: 3C3*(1/12)^3 + 3C2*(1/12)^2*(11/12)^1 the sum ofr P(7) + P(11) is what is wanted...

  17. anonymous
    • 5 years ago
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    sorry, your solution is wrong... probability that 7 or 11 appear on one throw of a pair of dice = 1/6 + 2/36 = 8/36 = 2/9 P(AUB) = P(A) + P(B) here P(A n B) = 0

  18. anonymous
    • 5 years ago
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    i think you're right...i approached it from a different angle but i think i messed it up that way...thanks...

  19. anonymous
    • 5 years ago
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    ok....

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