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Math Help - Probability

  1. #1
    Len
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    Probability

    Past experience has shown that, on average, only 1 in 10 wells drilled hits oil. Assume the events are independent.

    What is the probability that out of 5 drills, the company has 2 that hit oil?

    What is the probability that the second hit was on the fifth drill.

    Some help would be appreciated, thanks.
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    Quote Originally Posted by Len View Post
    Past experience has shown that, on average, only 1 in 10 wells drilled hits oil. Assume the events are independent.

    What is the probability that out of 5 drills, the company has 2 that hit oil?
    The probability of a hit is \frac{1}{10} and a miss is \frac{9}{10}, so if we want two hits, then we get \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3 =


    Quote Originally Posted by Len View Post
    What is the probability that the second hit was on the fifth drill.

    Some help would be appreciated, thanks.

    This one is a Negative Binomial Probability Distribution, which is defined as:

    \binom{y-1}{r-1}p^r(1-p)^{y-r} therefore we have:

    y=5, r=2, p=\tfrac{1}{10}

    then:

    \binom{5-1}{2-1}\left(\frac{1}{10}\right)^2\left( 1- \frac{1}{10} \right)^{5-2}=
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    Quote Originally Posted by Len View Post
    Past experience has shown that, on average, only 1 in 10 wells drilled hits oil. Assume the events are independent.

    What is the probability that out of 5 drills, the company has 2 that hit oil?

    [snip]
    Quote Originally Posted by lllll View Post
    The probability of a hit is \frac{1}{10} and a miss is \frac{9}{10}, so if we want two hits, then we get \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3 =

    [snip]
    Small correction: It will be {\color{red}{5 \choose 2}} \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3 \, ....
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    Quote Originally Posted by Len View Post
    Past experience has shown that, on average, only 1 in 10 wells drilled hits oil. Assume the events are independent.

    [snip]

    What is the probability that the second hit was on the fifth drill.

    Some help would be appreciated, thanks.
    This one can also be done as follows:

    Pr(exactly 1 hit in first four drills) x Pr (hit on fifth drill), where:

    Pr(exactly 1 hit in first four drills) = {4 \choose 1} \left(\frac{1}{10}\right)^1 \left(\frac{9}{10}\right)^3.

    Pr (hit on fifth drill) = 1/10.
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