Consider a game that consists of dealing out a hand of two random cards from a deck of four cards.The deck contains the Ace of Spades (As) ,the Ace of Hearts (Ah) ,the King of Spades (Ks) and the 9 of Hearts (9h) .Aces count as 1 or 11.Kings count as 10.You are interested in the total count of the two cards,with a maximum count of 21 (that is,AsAh = 12) .Let X be the sum of the two cards.Find the probability model for X.

A) $1/41/41/41/4\begin{array} { l | l c r r } \mathrm { X } & 1 & 9 & 10 & 11 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4\end{array}$
B) $1/61/61/31/3\begin{array} { l | l r r r } \mathrm { X } & 12 & 19 & 20 & 21 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 6 & 1 / 6 & 1 / 3 & 1 / 3\end{array}$
C) $1/51/51/51/51/5\begin{array} { l | l r r r r } \mathrm { X } & 12 & 19 & 20 & 21 & 22 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 5 & 1 / 5 & 1 / 5 & 1 / 5 & 1 / 5\end{array}$
D) $1/121/121/61/31/3\begin{array} { l | l r r r r } \mathrm { X } & 2 & 12 & 19 & 20 & 21 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 12 & 1 / 12 & 1 / 6 & 1 / 3 & 1 / 3\end{array}$
E) $1/41/41/41/4\begin{array} { l | l r r r } \mathrm { X } & 12 & 19 & 20 & 21 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4\end{array}$

Question

Consider a game that consists of dealing out a hand of two random cards from a deck of four cards.The deck contains the Ace of Spades (As) ,the Ace of Hearts (Ah) ,the King of Spades (Ks) and the 9 of Hearts (9h) .Aces count as 1 or 11.Kings count as 10.You are interested in the total count of the two cards,with a maximum count of 21 (that is,AsAh = 12) .Let X be the sum of the two cards.Find the probability model for X.

A)

1/41/41/41/4\begin{array} { l | l c r r } \mathrm { X } & 1 & 9 & 10 & 11 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4\end{array}

B)

1/61/61/31/3\begin{array} { l | l r r r } \mathrm { X } & 12 & 19 & 20 & 21 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 6 & 1 / 6 & 1 / 3 & 1 / 3\end{array}

C)

1/51/51/51/51/5\begin{array} { l | l r r r r } \mathrm { X } & 12 & 19 & 20 & 21 & 22 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 5 & 1 / 5 & 1 / 5 & 1 / 5 & 1 / 5\end{array}

D)

1/121/121/61/31/3\begin{array} { l | l r r r r } \mathrm { X } & 2 & 12 & 19 & 20 & 21 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 12 & 1 / 12 & 1 / 6 & 1 / 3 & 1 / 3\end{array}

E)

1/41/41/41/4\begin{array} { l | l r r r } \mathrm { X } & 12 & 19 & 20 & 21 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4\end{array}

Dyteria Collins · Accepted Answer

Final Answer : B, Explanation :  Since there are only four cards, we can list all the possible outcomes and their corresponding sum:- AsAh = 12- AsKs = 12- AhKs = 11- As9h = 10- Ah9h = 10- Ks9h = 19Therefore, the probability model for X is:- P(X=12) = 2/6 = 1/3 (two possible outcomes: AsAh or AsKs)- P(X=11) = 1/6 (one possible outcome: AhKs)- P(X=10) = 2/6 = 1/3 (two possible outcomes: As9h or Ah9h)- P(X=19) = 1/6 (one possible outcome: Ks9h)Note that the sum of all probabilities is 1.

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