3. You have been given a 6-sided die which is not a fair die. Let X be the numbe

Question

3. You have been given a 6-sided die which is not a fair die. Let X be the number of dots on the uppermost face when you toss the die once. Suppose that you have been told that X has the probability mass function 0.2 fx (x)- 0.1 ifx=1,3,5 or 6, if = 2 or 4, 0 otherwise Also note that E(X) = 3.6 Your friend rolls a fair 6-sided die. Call Y the number your friend rolled. Your friend claims: "If you and your friend each toss your own die once, your roll will tend to be higher than your friend's roll" There are at least two obvious ways1 to turn this (rather vague) claim into a proper mathematical statement (1) Your average roll is higher than your friend's average roll, i.e., E(X) > E(Y); or (2) The probability of you rolling higher than your friend is greater than the probability of you rolling lower than your friend, i.e., P(X > Y) is larger than P(X Y), E(L) = P(X

Explanation / Answer

the pmf of X is

f(x)= 0.2 x=1,3,5,6

=0.1 x=2,4

=0 otherwise

so E[X]=0.2*(1+3+5+6)+0.1*(2+4)=3.6

E[Y]=3.5

so E[X]>E[Y] hence statement 1 is true

b) H=1 if X>Y

=0 otherwise

so E[H]=1*P[H=1]+0*P[H=0]=1*P[X>Y]=P[X>Y]

L=1 if X<Y

=0 otherwise

so E[L]=1*P[L=1]+0*P[L=0]=1*P[X<Y]=P[X<Y]  

c) E[H|X=1]=P[1>Y]=0 because the lowest value Y can take is 1

E[H|X=2]=P[2>Y]=P[Y=1]=1/6

E[H|X=3]=P[3>Y]=P[Y=1]+P[Y=2]=2/6

E[H|X=4]=P[4>Y]=P[y=1]+P[Y=2]+P[Y=3]=3/6=1/2

E[H|X=5]=P[5>Y]=P[Y=1]+P[Y=2]+P[Y=3]+P[Y=4]=4/6

E[H|X=6]=P[6>Y]=P[Y=1]+P[Y=2]+P[Y=3]+P[Y=4]+P[Y=5]=5/6

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