2 more math questions

Discussion in 'DAT Discussions' started by ATLATLATL, May 2, 2007.

  1. ATLATLATL

    2+ Year Member

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    if log2=A and log3=B then log six is?
    a+b
    a-b
    a*b
    a/b
    a^b

    answer is a+b
    i would've thought a*b

    oh and the other question is...

    with one fair die, find the probability of throwing two fours in five attempts?
     
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  3. gochi

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    I cant really explain this, but I know how it works, so let me illustrate.

    Log X + Log Y= Log(XY) likewise Log X - Log Y= Log(X/Y)

    These are just properties of LOGARITHMS.

    Q: if log2=A and log3=B then log six is?

    A: log 6= log2 + log3=log(2*3)=log6 which is the same as a=3;b=3--> ab.
     
  4. Lonely Sol

    Lonely Sol cowgoesmoo fan!
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    Would this a binomial probablity because probability of sucess equals 1-probability oif failure. The reason is because there is only two choices; rolling a 4 or not rolling a 4:

    Since you have 1/6 chances of throwing 4 and 5/6 chance of throwing anything but 4.

    Since, you want 2 4s in 5 attenpts, you do the following:


    (1/6*1/6*5/6*5/6*5/6 )*10 (I got 10 from 5C2)

    *plz correct me if I am wrong
     
  5. Streetwolf

    Streetwolf Ultra Senior Member
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    Want two 4s in 5 tries.

    5 choose 2 - picks the two throws that will be a 4.
    (1/6)^2 - odds of throwing two 4s.
    (5/6)^3 - odds of throwing three non-4s.

    10*(1/36)*(125/216) = 1250/7776 = (625/3888) = 0.16075

    So 16.075%.

    Correct, LS.
     
  6. Streetwolf

    Streetwolf Ultra Senior Member
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    Start with: X * Y
    = 10^(Log X) * 10^(Log Y) <-- 10^(Log X) = Log(10^X) = X
    = 10^(Log X + Log Y) <-- A^X * A^Y = A^(X+Y)

    So we have XY = 10^(Log X + Log Y).
    Take the Log of both sides:

    Log(XY) = Log(10^(Log X + Log Y)
    Log(XY) = Log X + Log Y

    Try it for Log(X/Y).
     

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