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Thread: Probability - independent trials

  1. #1
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    Probability - independent trials

    A missile hits it's target with probability 0.3. How many missiles should be fired so that there is atleast an 80 % probability of hitting a target ?

    (Atleast give a hint)
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  2. #2
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    Re: Probability - independent trials

    After multiple missiles, the outcomes you care about are "Target was hit" and "Target was not hit". Let's find the probability that after firing $\displaystyle n$ missiles, the target was not hit at all. That means every missile missed. The probability for that would be $\displaystyle (1-0.3)^n$ since the probability for each missile is independent. Since there are only two possible outcomes, the probability that the target was hit is therefore $\displaystyle 1 - (1-0.3)^n$. Since you want that probability to be greater than 80%, you set $\displaystyle 1-(1-0.3)^n > 0.8$. Solve for $\displaystyle n$.

    Edit: If the original post was supposed to show that the missile hits its target with probability 0.3%, then change the equation to $\displaystyle 1-(1-0.003)^n>0.8$.

    Solution (if the probability for hitting the target is 0.3=30%):

    $\displaystyle 0.2 > (0.7)^n$

    Take the natural log of both sides:

    $\displaystyle \ln(0.2) > n\ln(0.7)$

    Since $\displaystyle \ln(0.7)<0$, when we divide both sides by a negative number, it flips the inequality:

    $\displaystyle n > \dfrac{\ln(0.2)}{\ln(0.7)} \approx 4.5$, hence, you need at least 5 missiles.

    If the probability was 0.3% = 0.003, then change that to $\displaystyle n > \dfrac{\ln(0.2)}{\ln(0.997)} \approx 535.674$, so you need at least 536 missiles.
    Last edited by SlipEternal; Jun 23rd 2014 at 12:02 PM.
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