Question

giulia is playing a game with 6 cards: 4 kings, 1 queen, and 1 jack. she draws 1 card out of the stack of cards, replaces it, and then draws another card.
what is the probability that she will draw a king and then a jack, ( p(\text{king, then jack}) )?
( \frac{1}{36} )
( \frac{1}{24} )
( \frac{1}{9} )
( \frac{1}{6} )

Explanation:

Step1: Calculate the probability of drawing a king

The total number of cards is \(n = 6\), and the number of kings is \(m_{1}=4\). The probability of drawing a king, \(P(\text{king})\), is given by the formula \(P=\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}\). So \(P(\text{king})=\frac{4}{6}=\frac{2}{3}\).

Step2: Calculate the probability of drawing a jack

Since the card is replaced, the total number of cards remains \(n = 6\), and the number of jacks is \(m_{2} = 1\). The probability of drawing a jack, \(P(\text{jack})\), is \(P(\text{jack})=\frac{1}{6}\).

Step3: Calculate the probability of the combined event

For independent events \(A\) and \(B\) (drawing a king and then a jack with replacement are independent events), \(P(A\cap B)=P(A)\times P(B)\). Here \(A\) is drawing a king and \(B\) is drawing a jack. So \(P(\text{king, then jack})=\frac{2}{3}\times\frac{1}{6}=\frac{2}{18}=\frac{1}{9}\).

Answer:

\(\frac{1}{9}\) (the third option)