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Thread: [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.


    $\displaystyle n$ 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 $\displaystyle \frac{19}{21}$
    How many coins do you have?
    When $\displaystyle n$ coins are flipped, there are $\displaystyle 2^n$ possible outcomes.

    We are told that at least one of the coins was Heads.
    . . Then there are: .$\displaystyle 2^n - 1$ 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 $\displaystyle n$ outcomes with exactly 1 Head.
    Hence, there are: .$\displaystyle 2^n - n - 1$ outcomes with at least 2 Heads.


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

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

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

    There is no elementary method for solving this equation.

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

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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 $\displaystyle 2^n$ 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 $\displaystyle 2^n-1$ 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 $\displaystyle 2^n$ total and subtract off the number with one head or no heads: $\displaystyle 2^n- n- 1$. The probability that "given that there is at least one head, there are at least 2" is $\displaystyle \frac{2^n-n-1}{2^n-1}$ 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!), $\displaystyle \frac{2^2- 2-1}{2^n-1}= \frac{1}{3}$.
    If n= 3, $\displaystyle \frac{2^3- 3- 1}{2^3-1}= \frac{4}{7}$.
    If n= 4, $\displaystyle \frac{2^4- 4- 1}{2^4-1}= \frac{11}{15}$
    If n= 5, $\displaystyle \frac{2^5- 5- 1}{2^5-1}= \frac{26}{31}$
    if n= 6, $\displaystyle \frac{2^6- 6- 1}{2^6-1}= \frac{57}{63}= \frac{19}{21}$!
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