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Math Help - Game show

  1. #1
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    Game show

    A contestant on the game show called "A Diamond for your Wife" gets to select 5 diamonds from a box. The box contains 20 diamonds of which 8 are fake and 12 are real. Regardless of how many real diamonds the contestant has after his selection he can only take one home for his wife. A second contestant then gets to select from the remaining 15 diamonds from the box. But only gets to select one diamond.

    What is the probability that this second contestant selects a real diamond?

    I don't understand why the following reasoning is wrong:

    Let F be the event where the diamond picked was fake and let R be the event where the diamond picked was real.

    Thus, the first contestant could get 0F, 1F, 2F, 3F, 4F, 5F.

    If the first contestant gets 0F, then that means he took 5R, thus the second contestant has a probability of 7/15 of getting 1R as there are now 7 R left out of the 15 diamonds.

    Likewise if the first contestant gets 1F, then that means he took 4R, thus the second contestant has a probability of 8/15 of getting 1R.

    This process continues until the first contestant picks 5F in which case the probability of the second contestant has a probability of 12/15 of getting 1R.

    So the total probability is 7/15 + 8/15 + ... + 12/15 but clearly this is greater than 1, where did I go wrong?
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  2. #2
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    Re: Game show

    Hello, usagi_killer!

    You omitted the probabilities for the first contestant.


    A contestant on the game show called "A Diamond for your Wife" gets to select 5 diamonds from a box.
    The box contains 20 diamonds of which 8 are fake and 12 are real.
    Regardless of how many real diamonds the contestant has after his selection, he can only take one home for his wife.
    A second contestant then gets to select one diamond from the remaining 15 diamonds from the box.

    What is the probability that this second contestant selects a real diamond?

    There are: . \begin{Bmatrix}\text{8 Fake} \\ \text{12 Real} \end{Bmatrix}


    Contestant-1 selects 5 diamonds.

    There are: . {20\choose5} \,=\,15,\!504 possible outcomes.


    There are six possible scenarios:


    [1] Contestant-1 chooses 0 Fake and 5 Reals.

    . . \text{0F, 5R:}\;{12\choose5} \:=\:792\text{ ways} \quad\Rightarrow\quad  P(\text{0F. 5R}) \:=\:\frac{792}{15,504} \:=\:\frac{99}{1938}

    . . \text{Then contestant-2 chooses from }\begin{Bmatrix}\text{8 F}\\\text{7 R}\end{Bmatrix} \quad\Rightarrow\quad P(R) \:=\:\frac{7}{15}

    . . \text{Hence: }\:P(\text{Scenario-1}\,\wedge\,R) \:=\:\frac{99}{1938}\cdot\frac{7}{15} \;=\;\frac{693}{29,\!070}


    [2] Contestant-1 choose 1 Fake and 4 Reals.

    . . \text{1F, 4R: }\;{8\choose1}{12\choose4} :=\:2640 \quad\Rightarrow\quad {(\text 1F, 4R}) \:=\:\frac{2640}{15,504} \:=\:\frac{330}{1938}

    . . \text{Then contestant-2 chooses from }\begin{Bmatrix}\text{7 F} \\\text{8 R}\end{Bmatrix} \quad\Rightarrow\quad P(R) \:=\:\frac{8}{15}

    . . \text{Hence: }\:P(\text{Scenario-2}\,\wedge R) \:=\:\frac{330}{1938}\cdot\frac{8}{15} \:=\:\frac{2640}{29,\!070}


    . . . . . . and so on.


    Got it?
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  3. #3
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    Re: Game show

    Quote Originally Posted by usagi_killer View Post
    A contestant on the game show called "A Diamond for your Wife" gets to select 5 diamonds from a box. The box contains 20 diamonds of which 8 are fake and 12 are real. Regardless of how many real diamonds the contestant has after his selection he can only take one home for his wife. A second contestant then gets to select from the remaining 15 diamonds from the box. But only gets to select one diamond.
    What is the probability that this second contestant selects a real diamond?
    I will follow some of your notation.
    Let F_k denote the event that the first contestant draws k fake diamonds.
    Let D denote the event that the second contestant draws k a real diamonds.
    P(D) = \sum\limits_{k = 0}^5 {P\left( {D \cap {F_k}} \right)}  = \sum\limits_{k = 0}^5 {P\left( {D|{F_k}} \right)} P\left( {{F_k}} \right)

    As Soroban has shown you P\left( {{F_k}} \right)=\dfrac{\binom{8}{k}\binom{12}{5-k}}{\binom{20}{5}} and {P\left( {D|{F_k}} \right)}=\frac{7+k}{15}.

    If you do the calculations the answer is surprising.
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  4. #4
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    Re: Game show

    Ahh yes thank you very much both of you, very clear and concise explanations!
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