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Thread: Independence and Multiplication rule help

  1. #1
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    Independence and Multiplication rule help

    Hi guys, I'm back again with another question hopefully you guys can help me.

    [Question]
    We have a fuse box with 20 fuses, 5 of which are defective.

    If 2 fuses are selected at random, in succession, and without replacement, what is
    the probability that both fuses are defective?

    [SOLN]
    A = "1st fuse is defective", P(A) = 5/20 = 1/4
    B = "2nd fuse is defective", P(B) = 4/19

    P(A n B) = (1/4) * (4/19) = 1/19


    ? - My confusion has to do with independence, since you can't calculate P(B|A) without having P(A n B)
    and since the above solution only seems to be calculable if you know the events are independent. Since then P(B|A) = P(B).

    But it seems to me that the events would be dependent since there is no replacement. For instance if the first fuse chosen
    was not defective then wouldn't the probability of choosing a defective fuse on the 2nd try actually be 5/19?

    I guess then my actual question is how can I figure out whether or not the events are dependent if (A n B) is not known?
    I hope I am making some sort of sense.

    Thanks for your help guys, I sure need it.
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  2. #2
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    Re: Independence and Multiplication rule help

    The events certainly are dependent.

    The probability the first fuse is defective is $\dfrac {5}{20}=\dfrac 1 4$

    This leaves 4 defective fuses in the 19 total remaining.

    So the probability that the 2nd fuse is defective, given the fact that the first one is, is $\dfrac {4}{19}$

    And the probability of these two events

    $P[\mbox{fuse 1 and fuse 2 were defective}] = P[\mbox{fuse 2 is defective |fuse 1 was defective}]P[\mbox{fuse 1 was defective}] = \dfrac{4}{19} \dfrac {1}{4} = \dfrac {1}{19}$
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  3. #3
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    Re: Independence and Multiplication rule help

    Hi Romsek, thanks for your reply. I'm still confused though.

    P(B|A) = P(B), IF the events are independent
    P(B|A) != P(B), IF the events are dependent

    But in this solution we say they are dependent, but then calculate as if it were independent?
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  4. #4
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    Re: Independence and Multiplication rule help

    Quote Originally Posted by mathgex View Post
    Hi Romsek, thanks for your reply. I'm still confused though.

    P(B|A) = P(B), IF the events are independent
    P(B|A) != P(B), IF the events are dependent

    But in this solution we say they are dependent, but then calculate as if it were independent?
    there's nothing about independence in the solution.

    $P[A \cap B] = P[A | B]~P[B]$ regardless of whether $A$ and $B$ are independent or not.

    maybe a better way to look at it is

    $P[A \cap B] = P[A | B]~P[B]$

    if $A$ and $B$ are independent then $P[A | B] = P[A]$ and we get the well known

    $P[A \cap B] = P[A]~P[B]$ for independent events.
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  5. #5
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    Re: Independence and Multiplication rule help

    So are we are saying then that P(B) is not actually 4/19 but P(B|A) is 4/19..
    And we simply just aren't considering P(B) in this case?

    P(B), in my mind, is probability that 2nd fuse is defective regardless of what was chose first, I guess.
    P(B) = ?? and P(B|A) = 4/19


    Changing the solution to:

    A = "1st fuse is defective", P(A) = 5/20 = 1/4
    (B|A) = "2nd fuse is defective also", P(B|A) = 4/19

    P(A n B) = P(A) * P(B|A) = (1/4) * (4/19) = 1/19
    Last edited by mathgex; Oct 25th 2015 at 08:39 PM.
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  6. #6
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    Re: Independence and Multiplication rule help

    Quote Originally Posted by mathgex View Post
    So are we are saying then that P(B) is not actually 4/19 but P(B|A) is 4/19..
    And we simply just aren't considering P(B) in this case?
    Yes. Let's take a look at what P[B] would be.

    $P[B] = P[B | A]~P[A] + P[B | !A]~P[!A]$

    Just to be clear
    A = bad first fuse
    !A = good first fuse

    as before

    $P[B | A ] = \dfrac 4 {19}, ~~P[A]=\dfrac 1 4$

    however

    $P[B | !A ] = \dfrac 5 {19}$

    and $P[!A] = \dfrac {15}{20} = \dfrac 3 4$

    so $P[B] = \dfrac {4}{19} \cdot \dfrac {1}{4} + \dfrac {5}{19} \cdot \dfrac{3}{4} = \dfrac {1}{19}+\dfrac{15}{76} = \dfrac {19}{76}= \dfrac {1}{4}$


    P(B), in my mind, is probability that 2nd fuse is defective regardless of what was chose first, I guess.
    exactly
    Thanks from mathgex
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  7. #7
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    Re: Independence and Multiplication rule help

    Ahhh excellent, thank you so much for your help on this Romsek, clear as a bell now.
    You rock!
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