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Math Help - Coin Flip Probability

  1. #1
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    Coin Flip Probability

    Good Afternoon All,

    I am doing a review and I cannot seem to figure out 2 problems that I must know for my actual classwork....

    --> What is the probability of getting at least 3 tails when tossing a coin six times

    (I understand that get all tails would be 1/64 but when I try to substitute the prob for getting a tails it is 1/2 and I get all confused...)

    ----> A bus has 7 stops and 4 passengers. Given that each passenger is equally likely to get off at any stop, What is the probability that 2 will get off at the same time.

    Any help would be greatly appreciated!

    Antony
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  2. #2
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    Quote Originally Posted by noncentz View Post
    Good Afternoon All,

    I am doing a review and I cannot seem to figure out 2 problems that I must know for my actual classwork....

    --> What is the probability of getting at least 3 tails when tossing a coin six times

    (I understand that get all tails would be 1/64 but when I try to substitute the prob for getting a tails it is 1/2 and I get all confused...)

    ----> A bus has 7 stops and 4 passengers. Given that each passenger is equally likely to get off at any stop, What is the probability that 2 will get off at the same time.

    Any help would be greatly appreciated!

    Antony
    The first question deals with cumulative distribution function.

    P(\text{desired})=1-P(X \leq 2) = 1 - \left( \sum_{i=0}^2 \binom{n}{i}p^i(1-p)^{n-i} \right)

    = 1 - \binom{6}{0}p^0(1-p)^{6} - \binom{6}{1}p^1(1-p)^{5} - \binom{6}{2}p^2(1-p)^{4}

    where p = \frac{1}{2}.

    For the second problem, does it mean exactly two get off at the same stop, or at least two?
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  3. #3
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    ----> What is the probability of getting at least 3 tails when tossing a coin six times

    (I understand that get all tails would be 1/64 but when I try to substitute the prob for getting a tails it is 1/2 and I get all confused...)

    1st flip: .5 = 0 tails, .5 = 1 tails
    2nd flip: .25 = 0 tails, .5 = 1 tails, .25 = 2 tails
    3rd flip: .125 = 0 tails, .375 = 1 tails, .375 = 2 tails, .125 = 3 tails
    4th flip: .0625 = 0 tails, .25 = 1 tails, .375 = 2 tails, .25 = 3 tails, .0625 = 4 tails

    ...
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  4. #4
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    Thanks for the response good sir or madam,

    Sorry about that ...

    A bus has 7 stops and 4 passengers. Given that each passenger is equally likely to get off at any stop, What is the probability that 2 will get off at the same stop.

    --> I think the question is asking me the likelyhood of two passenger's getting off on the same stop.
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  5. #5
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    Quote Originally Posted by noncentz View Post
    Thanks for the response good sir or madam,

    Sorry about that ...

    A bus has 7 stops and 4 passengers. Given that each passenger is equally likely to get off at any stop, What is the probability that 2 will get off at the same stop.

    --> I think the question is asking me the likelyhood of two passenger's getting off on the same stop.
    Well you didn't answer my question. If three people get off at the same stop, does that count?

    Let's assume yes.

    Then we want 1 - (probability that they all get off at different stops).

    So there are \binom{7}{4} ways to pick the stops, and for each one, there are 4! ways to permute them. So overall,

    P(\text{desired})=1-P(\text{all have distinct stops}) = 1 - \binom{7}{4}(4!)\left(\frac{1}{7^4}\right)

    Note that P(7,4) = \binom{7}{4}(4!) where P(n,k) denotes the number of ordered k-subsets of {1,2,...,n}.
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  6. #6
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    Thank you guys I managed to get my work done and thas a huge help. Binomial distribution is under my belt now
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