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MathGroup Archive 2005

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Re: can Mathematica be useful for this?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56576] Re: can Mathematica be useful for this?
  • From: "Kezhao Zhang" <kezhao.zhang at gmail.com>
  • Date: Fri, 29 Apr 2005 03:20:50 -0400 (EDT)
  • References: <d4kks5$eck$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Here is another approach:

For each language, let's use 1 to indicate that the person can speaks
it and 0 that he cannot. So a binary vector of length 3 describes what
languages the person speaks. For example, {S=1, F=0, G=0}, or {1, 0, 0}
means that he can speak Spanish only. The probably corresponding to
such vector is denoted by p[S,F,G] (lower case p).  All the
probabilities listed in the problem are functions of p[S,F,G] (all 8 of
them). For example, probability that the person can speak S and F, P[S
and F]=p[1,1,0]+p[1,1,1] (note the difference between upper case P and
lower case p), etc.

We need to write P in terms of p and solve for p which is the answer we
need.

The probabilities listed in the problem are

prob={P[1,0,0],P[0,1,0],P[0,0,1],P[1,1,0],P[0,1,1],P[1,0,1],P[1,1,1]};

unknown=Outer[p,Sequence@@Table[{0,1},{3}]]//Flatten
{p[0,0,0],p[0,0,1],p[0,1,0],p[0,1,1],p[1,0,0],p[1,0,1],p[1,1,0],p[1,1,1]}

Now write each of them in terms of p:

convert[index_P]:=Plus@@Cases[unknown,index/.{P->p,0->_}]

p@@@convert[#]&/@prob

{p[1,0,0]+p[1,0,1]+p[1,1,0]+p[1,1,1],p[0,1,0]+p[0,1,1]+

p[1,1,0]+p[1,1,1],p[0,0,1]+p[0,1,1]+p[1,0,1]+p[1,1,1],p[1,1,0]+p[1,1,1],
    p[0,1,1]+p[1,1,1],p[1,0,1]+p[1,1,1],p[1,1,1]}

The equation to be solved:

eqn=Thread[%=={1/3,26/100,2/10,15/100,5/100,1/10,2/100}];

Plus the condition that all p's sum to 1:

Solve for p:

sol=Solve[Append[eqn, Total[unknown]==1],unknown]

{{p[0,0,0]->73/150,p[0,0,1]->7/100,p[0,1,0]->2/25,p[0,1,1]->3/100,p[1,
    0,0]->31/300,p[1,0,1]->2/25,p[1,1,0]->13/100,p[1,1,1]->1/50}}

Hence the probability that the person doesn't speak any is
p[0,0,0]=73/150, and the probability that he speaks one only is

p[0,0,1]+p[1,0,0]+p[0,1,0]/.sol

{19/75}

K. Zhang


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