2006 dse q31

[chhavigupta]

i didnt get the solution that u posted earlier
Q31> P (A winning the contest) = [(1/2)2 + (1/2)5 + (1/2)8 + (1/2)11 +…]
                                                   + [(1/2)4+ (1/2)7 + (1/2)10 + ……]
                                                   = 4/14 + 1/14
                                                   = 5/14

can u explain the steps how we solved..

[duck]

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[chhavi]

three players A ,B ,C take turns playing a game as follows. A and B play in the first round . The winner plays C in the second round, while loser sits out. The winner of the second round plays the person who was sitting out. The game continues in this fashion , with the winner of the current round playing the next round with the person who sits out in the current round .the game ends when a player wins twice in succession ,this player is the winner of the contest. For any of the rounds, assume that the two players playing the round each hav a probability 1/2 of winning the round , regardless of how the past rounds were won or lost.
 

Ques. The probability that A BECOMES WINNER OF THE CONTEST--
a 5/14
b 1/2
C 3/7
D 7/16

Plz answer..how we get the answer as written in the above post..

[s-2]

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[chhavi]

thanks... I have one more doubt...
Can u explain exactly whats is meant by anti symmetric relation and quasitransitivity??...

Secondly.,  one more ques..dse 2009
  mark which one of the following is quasitransitive...and why???
 A ..xPy, yPz, zPx
B ... xPy, yPz, zIx
C ... xPy, yIz, zIx
D...  yPx, yIz, xPz
Where P is strictly preferred
I is indifferent relation.


Thirdly.. I want to get good grasp of such concepts n want to practice more ques..can u suggest some book..??

[duck]

hi.. :)

Let, a relation R is defined on the set S.
Anti symmetry: for all x,y belongs to S: {(x,y) belongs to R and (y,x) belongs to R => x=y}
Quasi transitivity: for all x,y,z belongs to S: {(x,y) belongs to P and (y,z) belongs to P => (x,z) belongs to P}


For the question you asked , you can straight away rule out options A,B and D by using the definition of Quasi transitivity.
However, in option C,Quasi transitivity is satisfied trivially.
So, option C is the answer.

[Priyanka kothari]

Hi!

 few definition try to understand them :(a) Reflexivity: for every x belongs to S : (x; x) belongs to R
(b) Completeness:for every x; y belongs to S : x 6= y ) (x; y) belongs to R or (y; x) belongs to R
(c) Transitivity: for every x; y; z belongs to S : ((x; y) belongs to R and (y; z) belongs to R) ) (x; z) belongs to R
(d) Symmetry: for every x; y belongs to S : (x; y) belongs to R ) (y; x) belongs to R
(e) Anti-symmetry: for every x; y belongs to S : ((x; y) belongs to R and (y; x) belongs to R) ) x = y
(f) Asymmetry: for every x; y belongs to S : (x; y) 2 R ) (y; x) does not belong to R
(g) Negative transitivity: for every x; y; z belongs to S : ((x; y) does not belong to R and (y; z) does not belong to R) ) (x; z) does not belong to R
(h) Equivalence: Relation which is symmetric, reflexive and transitive.

 do following question and explain your ans as well
1)S = {1,2,3}
 a)A ={(1,1), (2,2).(3,1)} determine if set A is reflexive(R/NR), transitive(T/NT) and    symmetric(S/NS)
 b) A = {(1,1),(2,1),(2,2),(3,3),(3,1)} determine if set S is reflexive(R/NR), transitive(T/NT)  and  symmetric(S/NS)

2)determine if set S is reflexive(R/NR), transitive(T/NT) and  symmetric(S/NS)        {N=not}
if S is the set  of all pairs of real nos. s.t
a) x<=y
B)x<y
C) x< lyl  
d)x^2 + y^2 = 1
e)x^2 + y^2 <0
f) x^2 + x=y^2 +y

[s-2]

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[Priyanka kothari]

hi shreya

well attempted.

for part c), its not transitive eg : take x= 2 y = -3 z = 1  
try part d) again, its not reflexive
part e) equivalence, as you have mentioned its empty set.... so, by definition it is reflexive transitive as well as symmetric.

let me know if you want to practice more i will post few more problems.
Enjoy!!

[bs5.cam786]

hi chavvi can u pls tell me that which book has this type of topic and answers, i mean which books has solution of this question if u dont mind i m a new member and starting the self study so i really need help