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bedsaul12345
10.10.2019 •
Mathematics
You and a friend play a game where you each toss a balanced coin. if the upper faces on the coins are both tails, you win $1; if the faces are both heads, you win $6; if the coins do not match (one shows a head, the other a tail), you lose $3 (win (−$ calculate the mean and variance of y, your winnings on a single play of the game. note that e(y) > 0.
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Ответ:
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)=![\dfrac{1}{4}=0.25](/tpl/images/0308/3340/b34f3.png)
You win $6; if the faces are both heads
then P(Y=6)=P(HH)=![\dfrac{1}{4}=0.25](/tpl/images/0308/3340/b34f3.png)
You loose $3; if the faces do not match.
then P(Y=1)=P(TH, HT)=![\dfrac{2}{4}=0.50](/tpl/images/0308/3340/302a8.png)
The expected value to win : E(Y)=![\sum_{i=1}^{i=3} y_ip(y_1)](/tpl/images/0308/3340/e472a.png)
Hence, the mean of Y : E(Y)= $0.25
Variance =![E[Y^2]-E(Y)^2](/tpl/images/0308/3340/8f7b8.png)
Hence, variance of Y = $ 6.19
Ответ: