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Math Help - A simple fair coin problem

  1. #1
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    A simple fair coin problem

    A fair coin is flipped 10 times. What is the probability that the heads turn up 8 or more times (8,9,10)?

    My approach:

    Probability that H turn up 8 times:
    P(8) = {10 \choose 8} (0.5)^{8} (0.5)^{2}
    Probability that H turn up 9 times:
    P(9) = {10 \choose 9} (0.5)^{10}
    Probability that H turn up 10 times:
    P(10) = {10 \choose 10} (0.5)^{10}

    I'm not 100% sure what to do now. Do I simply add the individual probabilities to get the total?

    P(8,9,10) = P(8) + P(9) + P(10)

    Thanks!
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by director View Post
    A fair coin is flipped 10 times. What is the probability that the heads turn up 8 or more times (8,9,10)?

    My approach:

    Probability that H turn up 8 times:
    P(8) = {10 \choose 8} (0.5)^{8} (0.5)^{2}
    Probability that H turn up 9 times:
    P(9) = {10 \choose 9}(0.5)^{10}
    Probability that H turn up 10 times:
    P(10) = {10 \choose 10} (0.5)^{10}

    I'm not 100% sure what to do now. Do I simply add the individual probabilities to get the total?

    \color{red}{\text{Yes.}}

    P(8,9,10) = P(8) + P(9) + P(10)

    Thanks!
    aNon1
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