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Chapter 16: Problem 4

A single card is drawn at random from a shuffled deck. What is the probability that it is red? That it is the ace of hearts? That it is either a three or a five? That it is either an ace or red or both?

Short Answer

Expert verified
The probabilities are: red card: \( \frac{1}{2} \), ace of hearts: \( \frac{1}{52} \), three or five: \( \frac{2}{13} \), ace or red or both: \( \frac{7}{13} \).

Step by step solution

01

- Total number of cards in a deck

First, determine the total number of cards in a standard deck. A standard deck has 52 cards.
02

- Probability a card is red

There are 26 red cards in the deck (13 hearts and 13 diamonds). The probability that a drawn card is red is: \[\frac{26}{52} = \frac{1}{2}\]
03

- Probability a card is the ace of hearts

There is only one ace of hearts in the deck. The probability of drawing the ace of hearts is: \[\frac{1}{52}\]
04

- Probability a card is a three or a five

There are 4 threes and 4 fives in a deck, making 8 such cards overall. The probability of drawing either a three or a five is: \[\frac{8}{52} = \frac{2}{13}\]
05

- Probability a card is an ace, red, or both

There are 4 aces and 26 red cards in the deck, but this counts the 2 red aces twice (ace of hearts and ace of diamonds). Hence, we use the principle of inclusion and exclusion: \[P(\text{ace} \cup \text{red}) = P(\text{ace}) + P(\text{red}) - P(\text{red ace})\]Substituting the corresponding probabilities, we get: \[\frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}\]

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Key Concepts

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

Probability Calculation
Probability is the measure of the chance that a particular event will occur. To calculate the probability of an event, you use the formula:

\( P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)

This formula gives you the likelihood of drawing a particular card from a deck, selecting a specific outcome, or any other similar event.
For example, to find the probability of drawing a red card from a standard 52-card deck, identify the number of red cards (26) and divide it by the total number of cards (52). This yields:
\[ P(\text{Red card}) = \frac{26}{52} = \frac{1}{2} \]
Knowing how to calculate probability enables you to understand and predict outcomes in various scenarios.
Deck of Cards
A standard deck of cards consists of 52 cards divided into 4 suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards: Ace through King.
  • There are 2 red suits: hearts and diamonds
  • There are 2 black suits: clubs and spades
  • Each suit has 1 Ace, 1 King, 1 Queen, 1 Jack, and number cards from 2 to 10
This structure is essential when calculating probabilities involving cards. For example, knowing there are 13 hearts allows you to calculate the probability of drawing the ace of hearts. Since there is only one ace of hearts in the deck, the probability is:
\[ P(\text{Ace of hearts}) = \frac{1}{52} \]
Being familiar with the deck’s structure makes these calculations straightforward and intuitive.
Inclusion-Exclusion Principle
The inclusion-exclusion principle is a technique used to calculate the probability of the union of multiple events. When events overlap (like being an ace and being red), simple addition of their probabilities would double-count certain outcomes. The formula is:
\[ P(A \bigcup B) = P(A) + P(B) - P(A \bigcap B) \]
This approach corrects for the over-counted intersections.
For example, to find the probability of drawing a card that is either an ace or red or both, calculate each separately, sum them, and subtract the overlap where they intersect (red aces):
\[ P(\text{ace} \bigcup \text{red}) = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13} \]
This corrects the double count and gives you the accurate probability.
Basic Combinatorics
Combinatorics involves counting, arranging, and combining objects effectively. It is the foundation for calculating probabilities. Common methods include permutations and combinations. In card games, understanding the number of ways to draw certain cards (favorable outcomes) helps determine probabilities.

For instance, consider the calculation for the probability of drawing either a three or a five. There are 4 threes and 4 fives, giving 8 favorable outcomes. The total number of possible outcomes being 52 cards:
\[ \frac{8}{52} = \frac{2}{13} \]
Basic combinatorics allows you to break down complex problems into simpler, manageable counts, making probability calculations easier.

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Most popular questions from this chapter

What is the probability that you and a friend have different birthdays? (For simplicity, let a year have 365 days.) What is the probability that three people have three different birthdays? Show that the probability that \(n\) people have \(n\) different birthdays is $$ p=\left(1-\frac{1}{365}\right)\left(1-\frac{2}{365}\right)\left(1-\frac{3}{365}\right) \cdots\left(1-\frac{n-1}{365}\right) $$ Estimate this for \(n \notin 365\) by calculating in \(p\) [recall that \(\ln (1+x)\) is approximately \(x\) for \(x<1]\). Find the smallest (integral) \(n\) for which \(p<\frac{1}{2}\). Hence show that for a group of 23 people or more, the probability is greater than \(\frac{1}{2}\) that two of them have the same birthdit? (Try it with a group of friends or a list of people such as the presidents of the United States.)

Recall that two events \(A\) and \(B\) are called independent if \(p(A B)=p(A) \cdot p(B) .\) Similarly, two random variables \(x\) and \(y\) with probability functions \(f(x)\) and \(g(y)\) are called independent if the probability of \(x=x_{i}\) and \(y=y_{j}\) is \(f\left(x_{i}\right) \cdot g\left(y_{j}\right)\) for every pair of values of \(x\) and \(y\), that is, if the joint probability function for \(x, y\) is \(f(x) g(y)\). Show that if \(x\) anct \(y\) are independent, then the expectation or average of \(x y\) is \(E(x y)=E(x) \cdot E(y)=\mu_{x} \mu_{y}\).

A circular garden bed of radius \(1 \mathrm{~m}\) is to be planted so that \(N\) seeds are uniformly distributed oser the circular area. Then we can talk about the number \(n\) of seeds in some particular area \(A\), or we can call \(n / N\) the probability for any one particular seed to be in the area \(A\). Find the probability \(F(r)\) that a seed (that is, some particular seed) is within \(r\) of the center.

Show that the expectation of the sum of two random variables defined over the same sample space is the sum of the expectations. Hint: Let \(p_{1}, p_{2}, p_{3}, \cdots, P_{*}\) be the probabilitics associated with the \(n\) sample points; let \(x_{1}, x_{2}, \cdots, x_{n}\), and \(y_{1}, y_{2}, \cdots, y_{n}\), be the values of the random variables \(x\) and \(y\) for the \(n\) sample points. Write out \(E(x), E(y)\), and \(E(x+y)\).

Two cards are drawn from a shuffled deck. What is the probability that both are red? If ar least one is red, what is the probability that both are red? If one is a red ace, what is the probability that both are red?
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