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6 days ago

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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 (−$3)). Calculate the mean and variance of Y, your winnings on a single play of the game. Note that E(Y) > 0.

1 answer:

Zina [9.1K]6 days ago

6 0

Answer: The mean and variance of Y are $0.25 and $6.19 respectively.

Step-by-step explanation:

The scenario is as follows: You and a friend participate in a game involving tossing a fair coin.

The sample space for tossing two coins is {TT, HT, TH, HH}

Let Y represent the earnings from one round of the game.

If both faces are heads, you win $1; therefore, P(Y=1)=P(TT)= $\dfrac{1}{4}=0.25$

You win $6 if both faces are heads, so P(Y=6)=P(HH)= $\dfrac{1}{4}=0.25$

If the faces do not match, you lose $3 which means P(Y=1)=P(TH, HT)= $\dfrac{2}{4}=0.50$

To find the expected value to win: E(Y)= $\sum_{i=1}^{i=3} y_ip(y_1)$

$=1\times0.25+6\times0.25+(-3)\times0.50=0.25$

Thus, the mean of Y: E(Y)= $0.25

$E(Y^2)=\sum_{i=1}^{i=3} y_i^2p(y_i)\\\\=1^2\times0.25+6^1\times0.25+(-3)^2\times0.5\\\\=0.25+1.5+4.5=6.25$

Variance = $E[Y^2]-E(Y)^2$

$=6.25-(0.25)^2=6.25-0.0625=6.1875\approx6.19$

Therefore, variance of Y = $ 6.19

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