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Math Help - Probability question

  1. #1
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    Probability question

    Three cards are drawn from an ordinary pack of 52 cards, at random and without replacement. Cards drawn have the following values:

    Aces score 1 point,
    Tens, Jacks, Queens and Kings score 10 points and
    Cards from two to nine score as many points as the numbers they carry (i.e. twos score 2 points, threes score 3 points and so on)

    Find the probabilities that
    ii) all three cards are of different suits (I found this, it's 197/425)
    iii) total score of the three cards is more than 28 given that all three cards are of different suits.

    what is the answer to iii)? I'm pretty sure my method is right, but I got 160/2197 and the solution given is 112/2197. If you can get 112/2197, please share. Thanks!

    My method:

    I found P(scores are 10, 10, 9 and of diff suits) + P(scores are 10, 10, 10 and of diff suits) and divided this by P(cards are of different suits). I got 160/2197 but apparently the answer is 112/2197.

    Thanks for your help!
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  2. #2
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    Re: Probability question

    Hello, thesocialnetwork!

    Three cards are drawn from an ordinary pack of 52 cards, at random without replacement.
    Cards drawn have the following values:
    . . Aces score 1 point,
    . . Tens, Jacks, Queens and Kings score 10 points,
    . . Other number cards score as many points as their numbers.

    Find the probabilities that:

    ii) all three cards are of different suits (I found this, it's 197/425)

    iii) total score of the 3 cards is more than 28, given that all 3 cards are of different suits.

    What is the answer to iii)?
    I'm pretty sure my method is right, but I got 160/2197
    and the solution given is 112/2197.
    If you can get 112/2197, please share. .Thanks!

    I don't agree with any of the answers.
    (But that could be my fault.)


    (ii) The 3 cards are of different suits.

    The first card can be any card: . \frac{52}{52} = 1
    . . The second card must be of another suit: . \frac{39}{51}
    . . . . The third card must be of yet another suit: . \frac{26}{50}

    P(\text{different suits}) \:=\:1\cdot\frac{39}{51}\cdot\frac{26}{50} \:=\:\frac{169}{425}



    (iii) P(\text{29 or 30}\,|\,\text{diff.suits}) \;=\;\dfrac{P(\text{29 or 30}\,\wedge\,\text{diff.suits})}{P(\text{diff.suit  s})}

    The numerator has two cases:

    . . P(29\,\wedge\,\text{diff.suits}) \quad\Rightarrow\quad 9,\,10,\,10\text{ of different suits.}
    . . . . There are {3\choose2} orders.
    . . . . The first card can be any Nine: . \frac{4}{52}
    . . . . . The second is a Ten of another suit: . \frac{3}{51}
    . . . . . . The third is a Ten of yet another suit: . \frac{2}{50}
    . . Hence: . P(29\,\wedge\,\text{diff.suits}) \:=\:{3\choose2}\frac{4}{52}\cdot\frac{3}{51}\frac  {2}{50} \:=\:\dfrac{3}{5525}

    . . P(30\,\wedge\,\text{diff.suits}) \quad\Rightarrow\quad 10,10,10 .
    (Of course, the 10's are of different suits!)
    . . . . There are {4\choose3} choices.
    . . . . The first card can be any Ten: . \frac{4}{52}
    . . . . The second card can be any other Ten: . \frac{3}{51}
    . . . . The third card can be any other Ten: . \frac{2}{50}
    . . Hence: . P(30\,\wedge\,\text{diff.suits}) \;=\;{4\choose3}\cdot\frac{4}{52}\cdot\frac{3}{51}  \cdot\frac{2}{50} \:=\:\dfrac{4}{5525}

    The numerator is: . \frac{3}{5525} + \frac{4}{5525} \:=\:\dfrac{7}{5525}

    The denominator was determined in part (ii): . \dfrac{169}{425}


    Therefore: . P(\text{29 or 30}\, |\,\text{diff.suits}) \;=\;\dfrac{\frac{7}{5525}}{\frac{169}{425}} \;=\;\frac{7}{5525}\cdot\frac{425}{169} \;=\;\frac{7}{2197}
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