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Math Help - studying; i still don't understand how to calculate this-binomial distribution

  1. #1
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    studying; i still don't understand how to calculate this-binomial distribution

    P(X< 3 n=15, successes=.4)
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  2. #2
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    Quote Originally Posted by beetz View Post
    i still don't understand how to calculate this-binomial distribution: P(X< 3 n=15, successes=.4)
    Think of a string of letters 15 places long.
    “FFFFFFFFFFFFFFF” that would be a string of 15 failures and can happen in only with probability (0.6)^{15}. That is because P(success) =0.4 while P(failure) ={1-(0.4)}=0.6.

    “SFFFFFFFFFFFFFF” that would be a string of 14 failures and one success. It can happen in 15 ways (S can be in 15 places). P(X = 1) = 15\left( {.04} \right)\left( {0.6} \right)^{14} .

    “SSFFFFFFFFFFFFF” that would be a string of 13 failures and two successes. It can happen in combination of 15 taken 2 at a time ways. P(X = 2) = {{15} \choose 2}\left( {.04} \right)^2 \left( {0.6} \right)^{13}

    P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
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  3. #3
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    hmmm...
    so, 0 + .0047 + 105 ( 15*14/2)* (.4)^2(.6)^13
    --- 0 + .0047 + .0219= .0266


    doesn't seem right....did i calculate wrong?

    do you know how to do this on a ti using the binom. functions?

    thanks.
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  4. #4
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    That is correct. I get 0.027114000777216
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  5. #5
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    oh, ok. sweeeeeeet.


    thanks for you help. i greatly appreciate it! :]
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  6. #6
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    now, if the problem were, say, less than 10, it would be the same concept, but for x=2, 15/2 (.4)^2 (.6)^13 and x=3 15/3 (.4)^3(.6)^11 etc. all the way to x=9?
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