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Two cards are drawn from a standard 52 -card playing deck. What is the probability that the draw will yield an ace and a face card?

Short Answer

Expert verified
The probability of drawing an ace and a face card is \( \frac{12}{221} \).

Step by step solution

01

Understand the Problem

We need to determine the probability of drawing an ace and a face card (Jack, Queen, King) from a deck of 52 cards. A standard deck has 4 aces and 12 face cards.
02

Calculate Total Possible Outcomes

Since two cards are drawn, the total number of possible outcomes is calculated by choosing 2 cards out of 52, i.e., \(\binom{52}{2}\).
03

Calculate Successful Outcomes for Ace and Face Card

An ace and a face card can be drawn in two ways: 1 ace first and 1 face card second, or 1 face card first and 1 ace second. Calculate both combinations and add them. For each scenario, use the combination formula.
04

Calculate Probability

Use the formula for probability: \( P(A) = \frac{\text{Number of Successful Outcomes}}{\text{Total Possible Outcomes}} \). Plug in the values calculated from previous steps to find the probability.
05

Simplify and Conclude

After calculation, simplify the probability to find the final answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arranging, and combinations of objects. It often involves calculating the different ways to select objects from a set. In problems involving probability, combinatorics is key to determining possible outcomes.

The exercise requires us to apply combinatorics to determine the number of ways to draw two cards from a 52-card deck. We need to consider both the combinations of cards and the order, even though drawing an ace and a face card doesn't have to be in a specific order.

Below is the fundamental formula used in this problem:
  • Combination Formula: inom{n}{k} = \frac{n!}{k!(n-k)!} where \(n\) is the total number of items to choose from, and \(k\) is the number of items to choose.


By understanding and applying combinatorics, we can establish both total possible and successful outcomes, thus helping us navigate through probability efficiently.
Ace and Face Card
In a standard deck of cards, understanding the specific characteristics of certain cards is important for calculating probabilities. Aces and face cards are particular types of playing cards that are crucial for this probability exercise.

A standard deck of 52 cards includes:
  • 4 Aces – one for each suit (hearts, diamonds, clubs, spades)
  • 12 Face Cards – comprising 4 Jacks, 4 Queens, and 4 Kings, one for each suit.


To calculate the probability of drawing an ace and a face card, we need to recognize these subdivisions, since they directly influence the number of successful outcomes.

Each card type (Ace or Face) has an equal probability of appearing if selected randomly, but the specific focus on these card types manifests because of their pronunciation in games and card rules. Thus, learning to segregate the types of cards is fundamental in probability scenarios involving card games like the one in this problem.
Standard Deck of Cards
The term "standard deck of cards" refers to the widely recognized and traditional set of 52 playing cards, consisting of four suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards, ranging from ace to king. This uniform structure forms the basis of numerous card-based games and probabilistic exercises.

Each deck typically comprises:
  • 36 Number Cards (2-10, in four suits)
  • 4 Aces (one per suit)
  • 12 Face Cards (Jack, Queen, King in each suit)


Understanding the layout of a standard deck is essential when calculating probabilities or designing strategies for card games. The uniformity ensures that combinations and draw scenarios follow predictable rules, which can thus be analyzed using mathematical principles such as probability and combinatorics. In probability problems like drawing from a standard deck, knowing the structure allows for precise calculations of possible events.
Successful Outcomes
Identifying and calculating successful outcomes forms the core of probability problems. An outcome is deemed 'successful' if it matches the criteria set by the problem – in this case, drawing an ace and a face card.

When two cards are drawn from the deck, and we desire one to be an Ace and the other a Face Card, this constitutes a successful pairing.

To compute these, we consider:
  • Ways to choose 1 Ace from 4 available
  • Ways to choose 1 Face Card from 12 available

Combinatorics tells us that for each Ace drawn, we have multiple face cards to pair with, and vice versa. Hence, each scenario (Ace first or Face Card first) contributes to the count of successful outcomes.

Using combinatorial formulas, like inom{4}{1} for nailing an Ace and inom{12}{1} for a Face Card, determines the likelihood efficiently, leading to a calculated probability ratio which represents how likely an event is to happen in the given context.

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Most popular questions from this chapter

A population of voters contains 40\% Republicans and 60\% Democrats. It is reported that 30\% of the Republicans and \(70 \%\) of the Democrats favor an election issue. A person chosen at random from this population is found to favor the issue in question. Find the conditional probability that this person is a Democrat.

Americans can be quite suspicious, especially when it comes to government conspiracies. On the question of whether the U.S. Air Force has withheld proof of the existence of intelligent life on other planets, the proportions of Americans with varying opinions are given in the table. $$\begin{array}{lc} \hline \text { Opinion } & \text { Proportion } \\ \hline \text { Very likely } & .24 \\ \text { Somewhat likely } & .24 \\ \text { Unlikely } & .40 \\ \text { Other } & .12 \\ \hline \end{array}$$ Suppose that one American is selected and his or her opinion is recorded. a. What are the simple events for this experiment? b. Are the simple events that you gave in part (a) all equally likely? If not, what are the probabilities that should be assigned to each? c. What is the probability that the person selected finds it at least somewhat likely that the Air Force is withholding information about intelligent life on other planets?

In a game, a participant is given three attempts to hit a ball. On each try, she either scores a hit, \(H\), or a miss, \(M\). The game requires that the player must alternate which hand she uses in successive attempts. That is, if she makes her first attempt with her right hand, she must use her left hand for the second attempt and her right hand for the third. Her chance of scoring a hit with her right hand is .7 and with her left hand is \(.4 .\) Assume that the results of successive attempts are independent and that she wins the game if she scores at least two hits in a row. If she makes her first attempt with her right hand, what is the probability that she wins the game?

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