Answer:
a) PMF
[tex]P(Y=0) = 0.375\\\\P(Y=1)=0.5\\\\P(Y=2)=0.125[/tex]
b) Cumulative distribution function
[tex]P(Y\leq 0)=0.375\\\\P(Y\leq1)=0.875\\\\P(Y\leq2)=1[/tex]
c) E(Y)=0.75
E((Y-1)^2)=1
Step-by-step explanation:
To write the pmf of Y we have to study the possible outcomes of this game. As this is a discrete game with no many possible outcomes, we can write them all one by one.
1) Event: The face of the die is an odd number and the coin is head (Y=1).
Probability:
[tex]P_1=P(odd)*P(head)=0.5*0.5=0.25[/tex]
2) Event: The face of the die is an odd number and the coin is tail (Y=0).
Probability:
[tex]P_2=P(odd)*P(tail)=0.5*0.5=0.25[/tex]
3) Event: The face of the die is an even number and the coin is tossed head two times (Y=2).
Probability:
[tex]P_3=P(odd)*P(head)*P(head)=0.5*0.5*0.5=0.125[/tex]
4) Event: The face of the die is an even number and the coin is tossed tail two times (Y=0).
Probability:
[tex]P_4=P(odd)*P(tail)*P(tail)=0.5*0.5*0.5=0.125[/tex]
5) Event: The face of the die is an even number and the coin is tossed one time head and one time tail (Y=1). The probabilitiesare doubled because this could happen in two ways (first head or first tail).
Probability:
[tex]P_5=P(odd)*(P(tail)*P(head)+P(head)*P(tail))\\\\P_5=0.5*(0.5*0.5+0.5*0.5)=0.25[/tex]
Then we have the pmf of Y as:
[tex]P(Y=0) = P_2+P_4=0.25+0.125=0.375\\\\P(Y=1)=P_1+P_5=0.25+0.25=0.5\\\\P(Y=2)=P_3=0.125[/tex]
The cumulative distribution function of Y is
[tex]P(Y\leq 0)=P(Y=0)=0.375\\\\P(Y\leq1)=P(Y\leq0)+P(Y=1)=0.375+0.5=0.875\\\\P(Y\leq2)=P(Y\leq1)+P(Y=2)=0.875+0.125=1[/tex]
The expected value of Y is:
[tex]E(Y)=\sum p_iY_i=0.375*0+0.5*1+0.125*2=0+0.5+0.25=0.75[/tex]
The expected value of X=(Y-1)² is:
[tex]E(X)=\sum p_iX_i=\sum p_i(Y_i-1)^2\\\\E(X)=0.375*(0-1)^2+0.5*(1-1)^2+0.125*(2-1)^2\\\\E(X)=0.375*1+0.5*1+0.125*1=1[/tex]
