ISI 2014 Q26...Amit Sir plz help..!!!

[Granpa Simpson]

Pr{Min(X,Y)<=Z} = Pr(X<=Z,Y<=Z)=Pr(X<=Z)*Pr(Y<=Z) since the are independent,
so Pr{Min(X,Y)<=Z}=Z*Z=Z^2,
Proceeding in similar manner I am getting Pr{Max(X,Y)<=(1-Z)}=(1-Z)^2,
Now (1-Z)^2=Z^2,
using this I am getting Z=1/2 as the answer...Amit sir plz let me know where am I am doing the mistake..!!!!

[Amit Goyal]

Pr(max{X, Y} <= z)
= Pr(X <= z, Y <= z)
= Pr(X <= z)Pr(Y <= z)
= z^2

Pr(min{X, Y} <= 1-z)
= Pr(X <= 1-z, Y <= 1-z) + Pr(X > 1-z, Y <= 1-z) +Pr(X <= 1-z, Y > 1-z)
= Pr(X <= 1-z)Pr(Y <= 1-z) + Pr(X > 1-z)Pr(Y <= 1-z) +Pr(X <= 1-z)Pr(Y > 1-z)
= (1-z)^2 + z(1-z) + (1-z)z
= 1 - (z^2)

Using the information z^2 = 1- z^2, we get z = 1/sqrt{2}

[Granpa Simpson]

Thanx a lot sir..!!!

[Dreyfus]

Amit Sir,, why aren't we considering three cases in max function
I mean when X=Y ...the prob will be z^2
But y r we not considering when X<Y and Y<X
Please clarify sir

[Granpa Simpson]

In case of max{X,Y} both X and Y have to be less than Z, that is necessary and sufficient..however in Min{X,Y} either X or Y or can be cases for being less than 1-Z.

[Dreyfus]

But subhayu den also der can be three cases :
X=Y P(max(X,Y)<=z)=z^2
X>Y P(max(X,Y)<=z)= z ( since P(Y<=z)=1 as x is already less than z)
Y>X P(max(X,Y)<=z)=z ( since P(X<=z)=1 as y is less than z)
Then y these two cases are not being consider?

[Amit Goyal]

Vaibhav,

You are mistaken in the way you are looking at things. I suggest you redo the concepts from some good textbook.
Anyways, let us do the problem in the way you were trying to do it:
Pr(max{X, Y} <= z)
= Pr(max{X, Y} <= z, X = Y) + Pr(max{X, Y} <= z, X < Y) + Pr(max{X, Y} <= z, X > Y)

Clearly,
Pr(max{X, Y} <= z, X = Y) = 0
because Pr(X = Y) = 0.

Pr(max{X, Y} <= z, X < Y) = Pr(X < Y <= z) = 0.5 (z^2)

Pr(max{X, Y} <= z, X > Y) = Pr(Y < X <= z) = 0.5 (z^2)

Therefore, Pr(max{X, Y} <= z) = 0 +  0.5 (z^2) + 0.5 (z^2) = z^2

[Dreyfus]

Thank you Sir, i got my mistake