Answer: The mean and variance of Y is $0.25 and $6.19 respectively.
Step-by-step explanation:
Given : You and a friend play a game where you each toss a balanced coin.
sample space for tossing two coins : {TT, HT, TH, HH}
Let Y denotes the winnings on a single play of the game.
You win $1; if the faces are both heads
then P(Y=1)=P(TT)=[tex]\dfrac{1}{4}=0.25[/tex]
You win $6; if the faces are both heads
then P(Y=6)=P(HH)=[tex]\dfrac{1}{4}=0.25[/tex]
You loose $3; if the faces do not match.
then P(Y=1)=P(TH, HT)=[tex]\dfrac{2}{4}=0.50[/tex]
The expected value to win : E(Y)=[tex]\sum_{i=1}^{i=3} y_ip(y_1)[/tex]
[tex]=1\times0.25+6\times0.25+(-3)\times0.50=0.25[/tex]
Hence, the mean of Y : E(Y)= $0.25
[tex]E(Y^2)=\sum_{i=1}^{i=3} y_i^2p(y_i)\\\\=1^2\times0.25+6^1\times0.25+(-3)^2\times0.5\\\\=0.25+1.5+4.5=6.25[/tex]
Variance = [tex]E[Y^2]-E(Y)^2[/tex]
[tex]=6.25-(0.25)^2=6.25-0.0625=6.1875\approx6.19[/tex]
Hence, variance of Y = $ 6.19
