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 Post subject: probability question
PostPosted: Wed, 18 Apr 2012 06:07:10 UTC 
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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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 Post subject: Re: probability question
PostPosted: Wed, 18 Apr 2012 14:23:01 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 8112
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Hello, thesocialnetwork!

Quote:
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} \:=\:\dfrac{169}{425}



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

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 9: .\frac{4}{52}
. . . . The second is a 10 of another suit: .\frac{3}{51}
. . . . The third is a 10 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 10: .\frac{4}{52}
. . . . The second card can be any other 10: .\frac{3}{51}
. . . . The third card can be any other 10: .\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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 Post subject: Re: probability question
PostPosted: Wed, 18 Apr 2012 14:30:33 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6837
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
thesocialnetwork wrote:
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!


You didn't show how you get 160/2197, so I can't pin down where your mistake come from. But part (ii) shouldn't give you a prime numerator (the numerator should be divisible by 13^2, as a check).

Counting the ordered triples:
29 points: T-point Suit #1 (4 ways), T-point Suit #2 (4 ways), 9-point Suit #3 (1 way), 3 permutation of card values (T,T,9), 24 ways to choose the suits (#1,#2,#3), in total: 1152 triples.
30 points: T-point Suit #1 (4 ways), T-point Suit #2 (4 ways), T-point Suit #3 (4 ways), 1 permutation of card values (T,T,T), 24 ways to choose suits, in total: 1536 triples.
So there are 2688 triples good triples, out of 52*39*26, i.e. probability 112/2197 as given.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: probability question
PostPosted: Sat, 21 Apr 2012 14:15:16 UTC 
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Joined: Wed, 15 Feb 2012 07:19:33 UTC
Posts: 31
I noticed my mistake! Many thanks guys.


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