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Math Help - Chinese Roulette?

  1. #1
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    Chinese Roulette?

    My science instructor offered a free A on the next quiz for any team that can solve this problem, but the people in my team can't agree on the solution. We're all coming up with different answers.

    This is how the problem was originally presented in class--

    Prompt: If the Large Hadron Collider (LHC) in Switzerland kills six people every year, and China is home to one-sixth of the world's population, what is the probability that at least one of the six people killed by the LHC will be from China?

    Assumptions:
    1. The Large Hadron Collider (LHC) in Switzerland is responsible for the death of exactly six people every year.
    2. Geographical location, health, age, and gender have no bearing on the probability of an individual being killed by the LHC.
    3. There are exactly six billion people on the earth.
    4. Exactly one-sixth of the earth's population resides in China.

    One guy says that a Chinese person is guaranteed to die--but we all know he's wrong...1 isn't much of a probability. One guy says the probability is about 16.67% (a 1 in 6 chance). The third guy brought up permutations, but never developed an answer. I don't know what to think. Help?

    Mika, 9th Grade
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by hopemichelle189 View Post
    My science instructor offered a free A on the next quiz for any team that can solve this problem, but the people in my team can't agree on the solution. We're all coming up with different answers.

    This is how the problem was originally presented in class--

    Prompt: If the Large Hadron Collider (LHC) in Switzerland kills six people every year, and China is home to one-sixth of the world's population, what is the probability that at least one of the six people killed by the LHC will be from China?

    Assumptions:
    1. The Large Hadron Collider (LHC) in Switzerland is responsible for the death of exactly six people every year.
    2. Geographical location, health, age, and gender have no bearing on the probability of an individual being killed by the LHC.
    3. There are exactly six billion people on the earth.
    4. Exactly one-sixth of the earth's population resides in China.

    One guy says that a Chinese person is guaranteed to die--but we all know he's wrong...1 isn't much of a probability. One guy says the probability is about 16.67% (a 1 in 6 chance). The third guy brought up permutations, but never developed an answer. I don't know what to think. Help?

    Mika, 9th Grade
    The probability that a single random person who dies is Chinese is 1/6, and we may assume that the world population is large enough that we need not worry about the imapact of this single death on the statistics of a small number of other deaths. So the probability that the second is Chinese is also 1/6, and so on.

    Similarly the probability that the deaths are not of Chinese is 5/6, and as these are independent we multiply these probabilities together to get:

    P(no Chinese amoung 6 random deaths)=(5/6)^6

    So:

    P(at least one Chinese in 6 random deaths)=1-P(no Chinese amound 6 random deaths)

    RonL
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