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August 2

For those that enjoys probability puzzles

An initially non pregnant woman has three boyfriends (A,B and C). In a night of utter drunkenness, she went with one of them back to her room and had unprotected sex. The very next day , all three of her boyfriends where killed when their jet plane crashed into the twin towers in New York. None of their bodies were recovered. Each boyfriend has a style in lovemaking, Boyfriend A will kiss her one third of the time and use a condom half the time. Boyfriend B will kiss her half the time and use a condom one third of the time. Boyfriend C will kiss her definitely and never use a condom. Given that the woman is now pregnant from that night's sexual encounter and she remember being kissed. The kissing and using condom are events that are independent for boyfriend A and B. What is the probability for each of the boyfriend of being the father of her child? 220.239.51.150 (talk) 10:32, 2 August 2013 (UTC)[reply]

Insufficient information. Assuming uniform a priori distribution, 1/9, 2/9 and 2/3, respectively.--208.169.64.170 (talk) 12:12, 2 August 2013 (UTC)[reply]

Odds for A is 1/3×1/2= 1/6. Odds for B is 1/2×2/3=1/3. Odds for C is 1×1=1. So the credibilities are as 1:2:6 in accordance with the previous user. The result reflects the information given, which is therefore sufficient. Uniform prior distribution follows from the principle of insufficient reason. Bo Jacoby (talk) 14:12, 2 August 2013 (UTC).[reply]

My answer using Mathematica
kiss["A"]=1/3;
kiss["B"]=1/2;
kiss["C"]=1;
nocondom["A"]=1/2;
nocondom["B"]=1/3;
nocondom["C"]=1;

Next we do a Sum of person X where X={A,B,C}

probkissnocondom=Plus@@Map[1/3*kiss[#]*nocondom[#]&,{"A","B","C"}]
4/9

prob[x_]:=1/3*kiss[x]*nocondom[x]/probkissnocondom

Map[ {#,prob[#]} & , {"A","B","C"} ]
{ {A,1/8} , {B,1/8} , {C,3/4} }
220.239.51.150 (talk) 11:48, 3 August 2013 (UTC)[reply]

The probability of B not using a condom is 2/3, not 1/3.--63.136.113.135 (talk) 11:55, 3 August 2013 (UTC)[reply]

You are right. Re runing the mathematica code again, gives the result { {"A", 1/9}, {"B", 2/9}, {"C", 2/3} } 220.239.51.150 (talk) 03:46, 4 August 2013 (UTC)[reply]

orbifold questions

1) Given a manifold and a group acting on it, is the process of making an orbifold out of them a functor? Like, does the manifolds fundametal group nicely transform into the orbifolds fundamental group?

2) Is a projective space, gotten from taking the orbits of nonzero scalar matrices as the points(the orbits are lines thru origin, deleting origin) an orbifold, at least under the string theory definition of orbifold?

3) More generally, in the mathematical dfinition and/or int the string theory definition, can an orbifold be nontrivially a manifold? (I mean not just taking the trivial group and acting on the manifold with that)Thanks, Rich76.218.104.120 (talk) 12:44, 2 August 2013 (UTC)[reply]