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Math Help - A simple propability question

  1. #1
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    A simple propability question

    Two learners are due to take an exam, The probability that Andy will pass is 3/5 and the probability that Brenda will fail is 4/7

    first i need to calculate the probability that both will fail

    then i need the probability that only one will pass

    and..

    If brenda fails, the probability that she will pass next time is 2/5
    a) calculate the probability that brenda fails again
    b) calculate the probability brenda passes having taken the exam for the second time.

    Any help is much appreciated!
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  2. #2
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    Quote Originally Posted by jimmybob View Post
    Two learners are due to take an exam, The probability that Andy will pass is 3/5 and the probability that Brenda will fail is 4/7

    first i need to calculate the probability that both will fail
    First will fail is 1-3/5=2/5
    Second will fail is 1-4/7=3/7
    Thus, we multiply the probabilities,
    (2/5)(3/7)=6/35
    then i need the probability that only one will pass
    2 possibilities...
    1)First person fails and second person passes.
    2)First person passes and first person fails.
    The probability in #1 is:
    (2/5)(4/7)=8/35
    The probability in #2 is:
    (3/5)(3/7)=9/35
    This is an OR question. Meaning #1 OR #2 thus we add,
    8/35+9/35=17/35
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  3. #3
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    [soze=3]Hello, jimmybob![/size]

    Andy and Brenda take an exam,
    The probability that Andy passes is \frac{3}{5} and the probability that Brenda fails is \frac{4}{7}.

    (1) Find the probability that both will fail

    (2) Find the probability that only one will pass

    We have these individual probabilities:
    . . \begin{array}{cc}P(\text{Andy passes}) = \frac{3}{5} & P(\text{Andy fails}) = \frac{2}{5} \\ P(\text{Brenda passes}) = \frac{3}{7} & P(\text{Brenda fails}) = \frac{4}{7} \end{array}


    (1) P(\text{both fail}) \:=\:P(\text{Andy fails and Brenda fails}) \:=\:\frac{2}{5}\cdot\frac{4}{7}\:=\:\boxed{\frac{  8}{35}}


    (2) If exactly one of them passes, there are two cases:

    . . Andy passes and Brenda fails: . P(A^+\text{ and }B^-) \:=\:\frac{3}{5}\cdot\frac{4}{7} \:=\:\frac{12}{35}

    . . Andy fails and Brenda passes: . P(A^-\text{ and }B^+) \:=\:\frac{2}{5}\cdot\frac{3}{7}\:=\:\frac{6}{35}

    . . Hence: . P(\text{one passes}) \:=\:\frac{12}{35} + \frac{6}{35}\:=\:\boxed{\frac{18}{35}}

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  4. #4
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    Quote Originally Posted by jimmybob View Post
    The probability that Andy will pass is 3/5 and
    the probability that Brenda will fail is 4/7.
    The probability that both will fail: \left( {1 - \frac{3}{5}} \right)\left( {\frac{4}{7}} \right).

    The probability that only one will pass: \left( {\frac{3}{5}} \right)\left( {\frac{4}{7}} \right) + \left( {\frac{2}{5}} \right)\left( {\frac{3}{7}} \right).
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