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Math Help - Conditional Probabilities

  1. #1
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    Conditional Probabilities

    I have a question on a review that I just cannot crack.
    It goes as follows:
    "Event A occurs with probability .4. The conditional probability that A occurs given B occurs is .5, while the probability that A occurs given that B does not occur is .2. What is the conditional Probability that B occurs given A occurs."

    So:
    P(A) = .4
    P(A|B)=.5
    P(A|Not B) = .2.

    I've tried multiplying things out. I just can't get it .
    Can anyone help?
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  2. #2
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    P(A)=P(B)\cdot P(A|B) + P(\bar{B})\cdot P(A|\bar{B})
    0.4=P(B)\cdot 0.5 + P(\bar{B})\cdot 0.2
    0.4=0.5\cdot P(B) + 0.2 - 0.2\cdot P(B)
    so P(B)=\frac{2}{3}

    P(B|A)=\frac{P(A\cap B)}{P(B)}
    so from P(A|B)=\frac{P(A\cap B)}{P(A)}=0.5We know that P(A\cap B)=0.5\cdot 0.4 = 0.2so P(B|A)=\frac{0.2}{2/3}=0.3
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  3. #3
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    The answer that the packet says is correct is "5/8." I don't know why though. T_T
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  4. #4
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    Quote Originally Posted by tylerm874 View Post
    The answer that the packet says is correct is "5/8."
    It appears that the packet's supplied answer is incorrect.
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  5. #5
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    The thing is, that answer isn't even an option on the packet ....

    And I haven't learned to do the problem that way, so I'm still confused.
    Thanks for helping though.
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  6. #6
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    Perhaps I should have used Bayes Rule, sorry its been a little while since i did some stat, damn summer holidays.. Although this still doesn't help you?
     <br /> <br />
P(B|A)=\frac{P(B)\cdot P(A|B)}{P(A)}=\frac{2/3\cdot 0.5}{0.4}=\frac{5}{6}<br />
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  7. #7
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    Not an answer option either.
    Thanks a lot though :/ .
    Unfortunately my teacher hasn't taught us Bayes Rule.
    She taught us the basic addition and multiplication rules, and taught us conditional earlier this week.
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  8. #8
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    Quote Originally Posted by tylerm874 View Post
    I have a question on a review that I just cannot crack.
    It goes as follows:
    "Event A occurs with probability .4. The conditional probability that A occurs given B occurs is .5, while the probability that A occurs given that B does not occur is .2. What is the conditional Probability that B occurs given A occurs."

    So:
    P(A) = .4
    P(A|B)=.5
    P(A|Not B) = .2.

    I've tried multiplying things out. I just can't get it .
    Can anyone help?
    Construct a Karnaugh Table:

    \begin{tabular}{l | c | c | c} & A & A' & \\ \hline B & a & b & a + b\\ B' & c & d & c + d \\ \hline & 0.4 & 0.6 & 1 \\ \end{tabular}

    From the given data you have:

    Pr(A | B) = 1/2 => a/(a + b) = 1/2 => a = b .... (1)

    Pr(A | B') = 1/5 => c/(c + d) = 1/5 => d = 4c .... (2)

    a + c = 2/5 ..... (3)

    b + d = 3/5 .... (4)

    Solve equations (1), (2), (3) and (4) simultaneously for a, b, c and d and complete the above table. The answer to your question is Pr(B | A) = a/0.4 = 5a/2.

    I get Pr(B | A) = 5/6.
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