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PostPosted: Wed, 9 May 2012 05:33:01 UTC 
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Joined: Sun, 6 May 2012 20:53:41 UTC
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Hello, I need some help with the following problem:

Exercise 2.15 of Hogg & Craig 5th edition

Five cards are drawn at random and without replacement from a bridge deck. Let the random variables X1, X2 and X3 denote, respectively, the number of spades, the number of hearts, and the number of diamonds that appear among the five cards.
A. Determine the joint p.d.f. of X1, X2 and X3.
B. Find the marginal probability density functions of X1, X2, and X3.
C. What is the joint conditional p.d.f. of X2 and X3, given that X1 = 3?

Thanks for any help provided,


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PostPosted: Wed, 9 May 2012 11:10:35 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
carlosochoa1985 wrote:
Hello, I need some help with the following problem:

Exercise 2.15 of Hogg & Craig 5th edition

Five cards are drawn at random and without replacement from a bridge deck. Let the random variables X1, X2 and X3 denote, respectively, the number of spades, the number of hearts, and the number of diamonds that appear among the five cards.
A. Determine the joint p.d.f. of X1, X2 and X3.
B. Find the marginal probability density functions of X1, X2, and X3.
C. What is the joint conditional p.d.f. of X2 and X3, given that X1 = 3?

Thanks for any help provided,


Where is your work?

_________________
\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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PostPosted: Thu, 10 May 2012 02:46:49 UTC 
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Math Cadet

Joined: Sun, 6 May 2012 20:53:41 UTC
Posts: 6
Hello,

I really don´t know where to start,,

thanks,


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PostPosted: Thu, 10 May 2012 04:08:33 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
carlosochoa1985 wrote:
Hello,

I really don´t know where to start,,

thanks,


Introduce the random variable X_4 for the number of clubs, then (X_1,X_2,X_3,X_4) can be modelled as ...

_________________
\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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PostPosted: Fri, 11 May 2012 05:50:59 UTC 
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Math Cadet

Joined: Sun, 6 May 2012 20:53:41 UTC
Posts: 6
Thanks again,

But I´m still stuck,

A.
$P\left(X_{1},X_{2},X_{3}\right)= \frac {\left({13C_X_{1}}\right)\left({13C_X_{2}}\right)\left({13C_X_{3}}\right)\left({13C_\left(5-X_{1}-X_{2}-X_{3}\right)}\right)}{{\left(52C_5}\right)}$ \\

B.
$P\left(X_{1}\right)= \sum_{X_{2}=0}^{5} \sum_{X_{3}=0}^{5} {P\left(X_{1},X_{2},X_{3}\right)}$ No idea how!!\\

C.
$P\left(X_{2},X_{3}|X_{1}=3\right)=\frac{P\left(X_{2},X_{3}\right)}{P\left(X_{1}=3\right)}$ Once I crack C., this one is easy\\

Could someone confirm me if A. is correct??,, and if it´s possible how to continue with B.???

Thanks,


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