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Math Help - Probability

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    Probability

    The probability that a missle will hit a target is 0.9. The commander wants to ensure that the probability of hitting the target is more than 97%.

    a) Find the least number of missiles that should be fired at the target.

    b) Can we ensure a 100% hit? Why?
    Last edited by Punch; February 9th 2010 at 02:51 AM.
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    Quote Originally Posted by Punch View Post
    The probability that a missle will hit a tgarget is 0.9. The commander wants to ensure that the probability of hitting the target is more than 97%.

    a) Find the least number of missiles that should be fired at the target.

    b) Can we ensure a 100% hit? Why?
    You want at least one missle to hit the target so if the probability all rockets will miss is P, the probability at least one will hit is 1- P. The probability that one missile will miss the target is .1 so the probability that all of n missiles will miss is (0.1)^n. You want to find n so that (0.1)^n< .03

    To ensure 100% you would need (0.1)^n= 0 and that is not true for any n.
    Last edited by Grandad; February 9th 2010 at 07:30 AM. Reason: Fixed LaTeX
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    Quote Originally Posted by HallsofIvy View Post
    You want at least one missle to hit the target so if the probability all rockets will miss is P, the probability at least one will hit is 1- P. The probability that one missile will miss the target is .1 so the probability that all of n missiles will miss is [tex](0.1)^n[tex]. You want to find n so that (0.1)^n< .03

    To ensure 100% you would need (0.1)^n= 0 and that is not true for any n.
    P(Rocket misses)=\frac{0.1}{1}
     =1

    97%(1)=0.97

    1-0.97=0.03

    P(N number of rocket missing)= (0.1)^n

    (0.1)^n<0.03

    n<1.522

    Therefore, least number of missiles to be launched is equals to 1.

    b) For a 100% hit, (0.1)^n=0, however no value is true for this equation. therefore, it is not possible.
    Last edited by Punch; February 9th 2010 at 03:08 AM.
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    Hello Punch
    Quote Originally Posted by Punch View Post
    ... (0.1)^n<0.03

    n<1.522

    ...
    You haven't shown your working here.

    And your answer is clearly not correct, is it?
    Quote Originally Posted by Punch View Post
    ...Therefore, least number of missiles to be launched is equals to 1.
    The answer should be, of course, n > 1.523. So n = 2.

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello PunchYou haven't shown your working here.

    And your answer is clearly not correct, is it?The answer should be, of course, n > 1.523. So n = 2.

    Grandad
    Hi, can you please show me the workings to get to n> 1.523?

    This is because I used logarithms to solve it. However, since this question falls under the elementary maths category, I am unable to solve it this way but I can't think of another way to solve it ...

    Also, I am unable to work out the answer given by you which is [tex]
    n > 1.523 by using logarithms.
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    Hello Punch
    Quote Originally Posted by Punch View Post
    Hi, can you please show me the workings to get to n> 1.523?

    This is because I used logarithms to solve it. However, since this question falls under the elementary maths category, I am unable to solve it this way but I can't think of another way to solve it ...

    Also, I am unable to work out the answer given by you which is [tex]
    n > 1.523 by using logarithms.
    Yes, you do use logarithms to solve this equation, but be careful with the direction of the inequality sign. You get:
    (0.1)^n < 0.03

    \Rightarrow n\log(0.1) < \log(0.03)
    So far, so good. But you can't then say:
    \Rightarrow n < \frac{\log(0.03)}{\log(0.1)}
    Can you see why not? It's because you have divided both sides of the inequality by \log(0.1) and \log(0.1) is negative. So you must reverse the direction of the inequality, and say:
    \Rightarrow n > \frac{\log(0.03)}{\log(0.1)}
    If you're not sure, think about this simple example:
    -2x < -10
    Does that mean (dividing both sides by -2) that:
    x < \frac{-10}{-2}

    \Rightarrow x < 5 ?
    No, of course it should be:
    x > 5
    Is that OK now?

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello PunchYes, you do use logarithms to solve this equation, but be careful with the direction of the inequality sign. You get:
    (0.1)^n < 0.03

    \Rightarrow n\log(0.1) < \log(0.03)
    So far, so good. But you can't then say:
    \Rightarrow n < \frac{\log(0.03)}{\log(0.1)}
    Can you see why not? It's because you have divided both sides of the inequality by \log(0.1) and \log(0.1) is negative. So you must reverse the direction of the inequality, and say:
    \Rightarrow n > \frac{\log(0.03)}{\log(0.1)}
    If you're not sure, think about this simple example:
    -2x < -10
    Does that mean (dividing both sides by -2) that:
    x < \frac{-10}{-2}

    \Rightarrow x < 5 ?
    No, of course it should be:
    x > 5
    Is that OK now?

    Grandad
    Thanks, I didn't know that... Is it then true that if I were to divide both side of an equality by a negative number, I would have to flip the equality sign?
    If it is, what about dividing just one side with a negative number?

    Also, is there an alternative other than using logarithms?
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  8. #8
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    Quote Originally Posted by Punch View Post
    Thanks, I didn't know that... Is it then true that if I were to divide both side of an equality by a negative number, I would have to flip the equality sign?
    Yes. If you multiply or divide both sides of an inequality by a negative number, you flip the inequality sign.
    If it is, what about dividing just one side with a negative number?
    Now you know that's not going to make sense! You never divide just one side of either an equation or an inequality by something.
    Also, is there an alternative other than using logarithms?
    You could simply use a calculator, and multiply 0.1 by itself until the result was less than 0.03. (In fact, of course, you only need do it once, because 0.1^2 = 0.01.) Or you could say:
    (0.1)^n<0.03

    \Rightarrow \left(\frac1{10}\right)^n<\frac{3}{100}

     \Rightarrow \left(\frac{10}{1}\right)^n>\frac{100}{3} (Notice we flip the inequality sign here, too.)

    \Rightarrow 10^n > 33.33...

    \Rightarrow the minimum value of n is 2.
    Grandad
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  9. #9
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    So, logarithms is much more efficient and the alternative is using trail and error, thank you very much grand dad.
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