Problem 17 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Two cards are drawn from a pack of 52 cards. What is the probability that one of them is a queen and the other is an ace? (a) \(2 / 663\) (b) \(2 / 13\) (c) \(4 / 663\) (d) \(8 / 663\)

Step-by-Step Solution

Answer

The probability is \( \frac{8}{663} \), option (d).

Step 1: Understand the problem

We need to find the probability that when two cards are drawn from a deck of 52 cards, one card is a queen and the other is an ace.

Step 2: Identify total possible outcomes

When two cards are drawn from a deck of 52 cards, the number of ways to choose 2 cards is given by the combination formula \( \binom{52}{2} \). Therefore, the total number of possible outcomes is \( \binom{52}{2} = \frac{52 \times 51}{2} = 1326 \).

Step 3: Identify favorable outcomes

Let's find the number of favorable outcomes where one card is a queen and the other is an ace. There are 4 queens and 4 aces in the deck. The number of ways to choose one queen out of 4 is \( \binom{4}{1} = 4 \) and similarly, one ace out of 4 is also \( \binom{4}{1} = 4 \). Thus, the number of favorable outcomes is \( 4 \times 4 = 16 \).

Step 4: Calculate the probability

The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes: \( \frac{16}{1326} \). To simplify, divide both the numerator and the denominator by 2: \( \frac{16}{1326} = \frac{8}{663} \).

Step 5: Choose the correct answer

The simplified probability \( \frac{8}{663} \) matches choice (d) in the provided options.

Key Concepts

Combination Formula

The concept of combinations is crucial in probability, especially when it comes to drawing cards or choosing elements from a group where the order does not matter.

The combination formula, often represented as \( \binom{n}{r} \), calculates the number of ways to choose \( r \) items from a set of \( n \) items. It is given by the formula:

• \( \binom{n}{r} = \frac{n!}{r! \, (n - r)!} \)

Here, \(!\) denotes factorial, which means the product of all positive integers up to a certain number. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

In the context of card games, this formula allows us to calculate the number of ways to draw two cards from a deck of 52. For this exercise, \( n = 52 \) (total number of cards) and \( r = 2 \) (the number of cards we want to draw). Plugging these values into the formula gives us:

• \( \binom{52}{2} = \frac{52 \times 51}{2} = 1326 \)

This represents all possible ways to draw any two cards from the deck.

Deck of 52 Cards

A standard deck of 52 cards is a staple in many card games and understanding its structure is vital for calculating probabilities like in the given exercise.

A typical deck is divided into four suits:

• Spades
• Hearts
• Diamonds
• Clubs

Each suit contains 13 cards, which includes numbers from 2 to 10, and face cards: Jack, Queen, King, and Ace. In this setup, each deck has:

• 4 Queens (one in each suit)
• 4 Aces (one in each suit)

When dealing with probability questions involving cards, it's important to remember the composition of the deck to accurately determine the number of available outcomes and favorable outcomes.

Favorable Outcomes

In probability, favorable outcomes refer to the specific scenarios that satisfy the condition of the problem we are solving. For example, drawing one queen and one ace.

In our exercise, we have to find the favorable outcomes for drawing two cards, one being a queen and the other an ace. With:

• 4 queens in total
• 4 aces in total

The number of ways to pick one queen from the four is \( \binom{4}{1} = 4 \). Similarly, the number of ways to choose one ace is also \( \binom{4}{1} = 4 \). The favorable outcomes, therefore, are calculated by multiplying these two:

• \( 4 \times 4 = 16 \)

These 16 combinations include every possible way to draw one queen and one ace from the deck. Comparing this number to the total possible outcomes determines the probability of our condition being met, which is \( \frac{8}{663} \) when simplified.