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Math Help - Probability Question

  1. #1
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    Probability Question

    An archer has two attempts to hit a target. The probability that his first arrow hits is 0.4 and the probability that the second arrow hits is 0.5 Given that the probability that he hits the target with both arrows is 0.25, find the probability that he misses the target with borrow arrows.

    I don't understand why the probability that he hits the target with both arrows is 0.25...
    0.5*0.4 = 0.2 - shouldn't that be the probability of hitting the target twice?



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  2. #2
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    e^(i*pi)'s Avatar
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    I might be overlooking something (in fact I probably am) but isn't the chance that he misses given by (1-0.6)(1-0.5) = 0.3

    ie: the chance he misses first time multiplied by the chance he misses second time
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Ah, there is conditional probability involved.

    The probability that the first arrow hits is 0.4, ok?
    How do you obtain the probability that the second arrow hit? You take:

    P(hit, hit) + P(not hit, hit)

    Reading: Probability 1st hits and 2nd hits or probability 1st misses but second hits.

    Have a probability tree drawn to help you.

    Can you try this out now?
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  4. #4
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    Hello, juliak!

    An archer has two attempts to hit a target.
    The probability that his first arrow hits is 0.4.
    And the probability that the second arrow hits is 0.5.
    Given that the probability that he hits the target with both arrows is 0.25,
    find the probability that he misses the target with both arrows.

    Why is the probability that he hits the target with both arrows 0.25 ?

    (0.5)(0.4) \,=\, 0.2 . Shouldn't that be the probability of hitting the target twice?

    Only if the two events are independent.

    It could be that the probablity of his second arrow
    depends on the result of the first arrow.

    We are expected to know this formula:

    . . P(\text{1st hit} \vee \text{2nd hit}) \;=\;\underbrace{P(\text{1st hit})}_{0.40} + \underbrace{P(\text{2nd hit})}_{0.50} - \underbrace{P(\text{1st hit} \wedge \text{2nd hit})}_{0.25}

    . . P(\text{1st hit} \vee \text{2nd hit}) \;=\;0.65


    So we have: the probability that hits with his 1st arrow
    . . or his 2nd arrow or both is 0.65


    Hence, the probability that he misses with both arrows is the "opposite".


    Therefore: . P(\text{both miss}) \;=\;1 - 0.65 \;=\;0.35

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