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Thread: Pr radar detecting an aircraft

  1. #1
    Mar 2009

    Pr radar detecting an aircraft

    Hi All,

    I think I have the right answer for this, but I'm not that confident on these questions..

    Three radar sets, operating independently, are set to tdetect any aircraft flying through a certain area. Each set has a probability of .02 of failing to detect a plane in its area. If an aircraft enters the area, what is the probability that it
    a goes undeteected?
    b is detected by all three radar sets?

    So I defined the events;
    $\displaystyle B_{1}$ - first fails
    $\displaystyle B_{2}$ - second fails
    $\displaystyle B_{3}$ - third fails.

    Then $\displaystyle P(a)=P(B_{1} \cup B_{2} \cup B_{3})$$\displaystyle =P(B_{1})+P(B_{2})+P(B_{3}) $$\displaystyle - \left[ P(B_{1} \cap B_{2} )+P(B_{1} \cap B_{3} )+P(B_{2} \cap B_{3} )\right] + \left[ P(B_{1} \cap B_{2} \cap B_{3})\right]$

    With $\displaystyle P(B_{1} \cap B_{2})=P(B_{2} \cap B_{3} )=P(B_{2} \cap B_{3} )=0.02 \cdot 0.02 = 0.0004$
    and $\displaystyle P(B_{1} \cap B_{2} \cap B_{3})=0.02 \cdot 0.02 \cdot 0.02 = 0.000008$
    So $\displaystyle P(a) = 0.02+0.02+0.02-\left[ 0.0004+0.0004+0.0004\right] + 0.000008=0.058808$

    And b detected by all three;
    $\displaystyle P(b)=P(\bar{B_{1}} \cap \bar{B_{2}} \cap \bar{B_{3}}=0.98 \cdot 0.98 \cdot 0.98=0.941192$

    Thanks for any help
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  2. #2
    Pim is offline
    Dec 2008
    The Netherlands
    b is correct. However, a is not.

    $\displaystyle P(B_{1} \cup B_{2} \cup B_{3}) $
    is the probability that any of the three radars does not detect it.
    The probability that none detect it, would be:
    $\displaystyle P(B_{1} \cap B_{2} \cap B_{3}) $
    As this means: 1 does not detect AND 2 does not detect AND 3 does not detect. And you worked that one out already.
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