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in a new card game, you start with a well-shu ed full deck and draw 3 cards without replacement. if you draw 3 hearts, you win $50. if you draw 3 black cards, you win $25. for any other draws, you win nothing. (a) create a probability model for the amount you win at this game, and nd the expected winnings. also compute the standard deviation of this distribution. (b) if the game costs $5 to play, what would be the expected value and standard deviation of the net pro t (or loss)? (hint: pro t



Answer :

a) The expected winning probability for a single game defined is: $3.59

b) The standard deviation of winning is: $10.11

a) If the game costs $5 to play, the expected winnings are: - $0.79

If the game costs $5 to play, the standard deviation of winning is $11.33.

Because the projected winning value is negative, the game should not be played with a -$5 playtime charge.

The total number of hearts in a deck is 13.

The probability of drawing three hearts is:

P(drawing 3 Hearts)

13 ÷ 52 × 12 ÷ 51 × 11 ÷ 50 = 0.0129

The probability of selecting three black cards is:

A deck of black cards has 26 cards.

P (drew three black cards):

26 ÷ 52 × 25 ÷ 51 × 24 ÷ 50 = 0.1176

Any other draw probability: P(3 hearts) + P(3 blacks) + P(any other draw) = 1.

858 ÷ 66300 + 7800 ÷ 66300 + P(any other draw) = 1

P(any other draw) = 57642 ÷ 66300

For only one game,

E(X) = Σ[ X × p(X)]

E(X) =Σ[(50 × 858 ÷ 66300) + (25 × 7800 ÷ 66300)+0]

E(X) = $3.588

b) The standard deviation is equal to √Var(X)

Var(X) = Σ[ X² × p(X)] - E(X)

Σ[ X² × p(X)] = Σ[(50² × 858 ÷ 66300) + (25² × 7800 ÷ 66300) + 0] = 105.88235

Var(X) = 105.88235 - 3.588 = 102.29435

Winning standard deviation = √102.29435 = $10.114

If the game cost $5 to play :

The total sum won if and only if:

Three sketched hearts = $50 - $5 = $45

Three blacks are drawn for a total of $25 - $5 = $20.

Any other combination drawn = $0 - $5 = -$5

The distribution is as follows:

E(X) = Σ[ X × p(X)]

E(X) =Σ[(50 × 858 ÷ 66300) + (25 × 7800 ÷ 66300) + (-5 × 57642 ÷ 66300)]

E(X) = - $0.7588

The winning standard deviation is:

The standard deviation is equal to √Var(X)

Var(X) = Σ[ X² × p(X)] - E(X)

Σ[ X²×p(X)] = Σ[(50² × 858 ÷ 66300) + (25² × 7800 ÷ 66300) + (-5² × 57642 ÷ 66300)] = 127.61764

Var(X) = 127.61764 - (-0.7588) = 128.37644

The winning standard deviation is:

Std(X) = √Var(X) = √128.37644 = $11.330

With a game cost of - $5, the predicted winnings for a single game are negative, hence you should avoid playing the game.

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