Question

q32.) jenna has a deck of 13 cards including 4 kings, 4 queens, 4 jacks, and 1 ace. jenna draws two cards from the deck. what is the probability that at least one of the cards jenna drew is a king? answer

Explanation:

Step1: Calculate the probability of no - king in both draws

The probability of not drawing a king in the first draw: There are 4 kings in a deck of 13 cards, so the number of non - king cards is \(13 - 4=9\). The probability of not drawing a king in the first draw \(P(\text{not K}_1)=\frac{9}{13}\).
After the first non - king draw, there are 12 cards left, and 8 non - king cards left. The probability of not drawing a king in the second draw given that no king was drawn in the first draw \(P(\text{not K}_2|\text{not K}_1)=\frac{8}{12}\).
By the multiplication rule of probability for dependent events, the probability of not drawing a king in both draws is \(P(\text{not K}_1\cap\text{not K}_2)=P(\text{not K}_1)\times P(\text{not K}_2|\text{not K}_1)=\frac{9}{13}\times\frac{8}{12}=\frac{9\times8}{13\times12}=\frac{72}{156}=\frac{6}{13}\).

Step2: Calculate the probability of at least one king

The probability of at least one king is the complement of the probability of no kings. Let \(P(X\geq1)\) be the probability of at least one king. Then \(P(X\geq1)=1 - P(X = 0)\).
Since \(P(X = 0)=\frac{6}{13}\), then \(P(X\geq1)=1-\frac{6}{13}=\frac{13 - 6}{13}=\frac{7}{13}\).

Answer:

\(\frac{7}{13}\)