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Math Help - [SOLVED] N coins are flipped, what is the probability?

  1. #1
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    [SOLVED] N coins are flipped, what is the probability?

    n coins are flipped, and you are given that at least 1 of them is heads. The probability that at least one other coin is heads is 19/21. How many coins do you have?
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  2. #2
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    Hello, mark1950!

    An unusual problem . . .
    I had to come up with an unusual solution.


    n coins are flipped, and you are given that at least 1 of them is heads.
    The probability that at least one other coin is heads is \frac{19}{21}
    How many coins do you have?
    When n coins are flipped, there are 2^n possible outcomes.

    We are told that at least one of the coins was Heads.
    . . Then there are: . 2^n - 1 outcomes.

    We eliminate the case of "no Heads" (all Tails).

    We want the probability that at least two coins are Heads.
    . . There is 1 outcomes with 0 Heads.
    . . There are n outcomes with exactly 1 Head.
    Hence, there are: . 2^n - n - 1 outcomes with at least 2 Heads.


    The probability of at least two Heads, given there is at least one Head is \frac{19}{21}

    . . \frac{2^n - n - 1}{2^n - 1} \;=\;\frac{19}{21} \quad\Rightarrow\quad 21\cdot2^n - 21n - 21 \;=\;19\cdot2^n - 19

    . . 2\cdot2^n - 21n \:=\:2 \quad\Rightarrow\quad 2^{n+1} \;=\;21n + 2

    There is no elementary method for solving this equation.

    By inspection, I found the solution: . n = 6

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  3. #3
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    Erm, I don't really get this step. If you don't mind, can you please explain in a more specific detail? Thanks.

    We want the probability that at least two coins are Heads.
    . . There is 1 outcomes with 0 Heads.
    . . There are n outcomes with exactly 1 Head.
    Hence, there are: .2^n - n - 1 outcomes with at least 2 Heads.

    And also, how do you do inspections to get n = 6?
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  4. #4
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    Quote Originally Posted by mark1950 View Post
    Erm, I don't really get this step. If you don't mind, can you please explain in a more specific detail? Thanks.

    We want the probability that at least two coins are Heads.
    . . There is 1 outcomes with 0 Heads.
    . . There are n outcomes with exactly 1 Head.
    Hence, there are: .2^n - n - 1 outcomes with at least 2 Heads.

    And also, how do you do inspections to get n = 6?
    If Soroban doesn't mind my jumping in here:
    With n coins there are 2^n possible outcomes: each coin can come up either heads or tails so each coin multiplies the possiblities by 1. Exactly one of those has "all tails" so 2^n-1 have "at least one head.

    There are exactly n outcomes with exactly one head because there are n positions that "head" can be in. To find the number of possible outcomes with 2 or more heads, start with 2^n total and subtract off the number with one head or no heads: 2^n- n- 1. The probability that "given that there is at least one head, there are at least 2" is \frac{2^n-n-1}{2^n-1} and that must be 19/21.

    Now just calculate a few values: if n= 2 (there must be at least two flips to get two heads!), \frac{2^2- 2-1}{2^n-1}= \frac{1}{3}.
    If n= 3, \frac{2^3- 3- 1}{2^3-1}= \frac{4}{7}.
    If n= 4, \frac{2^4- 4- 1}{2^4-1}= \frac{11}{15}
    If n= 5, \frac{2^5- 5- 1}{2^5-1}= \frac{26}{31}
    if n= 6, \frac{2^6- 6- 1}{2^6-1}= \frac{57}{63}= \frac{19}{21}!
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