Question

Let a card be selected from an ordinary deck of playing cards. The outcome \(c\) is one of these 52 cards. Let \(X(c)=4\) if \(c\) is an ace, let \(X(c)=3\) if \(c\) is a king, let \(X(c)=2\) if \(c\) is a queen, let \(X(c)=1\) if \(c\) is a jack, and let \(X(c)=0\) otherwise. Suppose that \(P\) assigns a probability of \(\frac{1}{52}\) to each outcome \(c .\) Describe the induced probability \(P_{X}(D)\) on the space \(\mathcal{D}=\\{0,1,2,3,4\\}\) of the random variable \(X\).

Short answer

\(P_X(0)=\frac{9}{13}\), \(P_X(1)=\frac{1}{13}\), \(P_X(2)=\frac{1}{13}\), \(P_X(3)=\frac{1}{13}\), and \(P_X(4)=\frac{1}{13}\)

Step-by-step solution

Analyzing the random variable \(X(c)\)

The function \(X(c)\) maps the outcomes of the card draw to the set \(\mathcal{D}=\{0,1,2,3,4\}\). More specifically, this function assigns a value of 4 to aces, a 3 to kings, a 2 to queens, a 1 to jacks and a 0 to all other cards.

Calculating the probabilities for each possible value of \(X(c)\)

The probabilities for each possible value of \(X(c)\) can be calculated as follows:- \(P_X(4)=P(\{c: X(c)=4\})=P(\{ace\})=\frac{1}{52}*4\) because there are 4 aces in a deck of cards.- \(P_X(3)=P(\{c: X(c)=3\})=P(\{king\})=\frac{1}{52}*4\) because there are 4 kings in a deck of cards.- \(P_X(2)=P(\{c: X(c)=2\})=P(\{queen\})=\frac{1}{52}*4\) because there are 4 queens in a deck of cards.- \(P_X(1)=P(\{c: X(c)=1\})=P(\{jack\})=\frac{1}{52}*4\) because there are 4 jacks in a deck of cards.- \(P_X(0)=P(\{c: X(c)=0\})=P(\{all\:other\:cards\})=\frac{1}{52}*(52-16)\) because there are 52-16=36 other cards in a deck of cards.

Presenting the final induced probability distribution

The induced probability distribution on the space \(\mathcal{D}\) of the random variable \(X\) is then given by the following equalities:- \(P_X(0)=\frac{36}{52}=\frac{9}{13}\)- \(P_X(1)=\frac{4}{52}=\frac{1}{13}\)- \(P_X(2)=\frac{4}{52}=\frac{1}{13}\)- \(P_X(3)=\frac{4}{52}=\frac{1}{13}\)- \(P_X(4)=\frac{4}{52}=\frac{1}{13}\)