Question

a standard deck of 52 playing cards contains four of each numbered card 2–10 and four each of aces, kings, queens, and jacks. two cards are chosen from the deck at random. which expression represents the probability of drawing a king and a queen? \\(\frac{(_{4}p_{1})(_{3}p_{1})}{_{52}p_{2}}\\) \\(\frac{(_{4}c_{1})(_{3}c_{1})}{_{52}c_{2}}\\) \\(\frac{(_{4}p_{1})(_{4}p_{1})}{_{52}p_{2}}\\) \\(\frac{(_{4}c_{1})(_{4}c_{1})}{_{52}c_{2}}\\)

Explanation:

Step1: Determine if order matters

When drawing a king and a queen, the order (king then queen or queen then king) doesn't matter for the event of having one king and one queen. So we use combinations (since combinations are for unordered selections) instead of permutations (which are for ordered selections).

Step2: Calculate number of ways to choose a king

There are 4 kings, and we want to choose 1. The number of ways to do this is \(_{4}C_{1}\) (using the combination formula \(_{n}C_{k}=\frac{n!}{k!(n - k)!}\), here \(n = 4\), \(k=1\), so \(_{4}C_{1}=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4\)).

Step3: Calculate number of ways to choose a queen

There are 4 queens, and we want to choose 1. The number of ways to do this is \(_{4}C_{1}\) (similarly, \(n = 4\), \(k = 1\), so \(_{4}C_{1}=4\)).

Step4: Calculate total number of ways to choose 2 cards from 52

The total number of ways to choose 2 cards from 52 (unordered) is \(_{52}C_{2}\) (using the combination formula \(_{n}C_{k}=\frac{n!}{k!(n - k)!}\), here \(n = 52\), \(k = 2\), so \(_{52}C_{2}=\frac{52!}{2!(52 - 2)!}=\frac{52\times51}{2\times1}\)).

Step5: Form the probability expression

The probability of an event is the number of favorable outcomes divided by the number of total outcomes. The number of favorable outcomes (drawing one king and one queen) is the product of the number of ways to choose a king and the number of ways to choose a queen, i.e., \((_{4}C_{1})(_{4}C_{1})\). The number of total outcomes is \(_{52}C_{2}\). So the probability expression is \(\frac{(_{4}C_{1})(_{4}C_{1})}{_{52}C_{2}}\).

Answer:

\(\boldsymbol{\frac{(_{4}C_{1})(_{4}C_{1})}{_{52}C_{2}}}\) (the last option: \(\frac{(_{4}C_{1})(_{4}C_{1})}{_{52}C_{2}}\))