/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 From a pack of fifty-two cards a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 9

From a pack of fifty-two cards a card is withdrawn at random and not replaced. A second card is then drawn. What is the probability that the first card is an ace and the second card a king? 10 Two ordinary six-faced dice are tossed. Write down the sample space of all possible combinations of values. What is the probability that the two values are the same? What is the probability that they differ by at most one?

Short Answer

Expert verified
The probability of drawing an ace first and a king second is \(\frac{4}{663}\). The probability of rolling the same value is \(\frac{1}{6}\), and the probability of the values differing by at most 1 is \(\frac{11}{36}\).

Step by step solution

01

Calculate Probability of Drawing an Ace First

A standard deck contains 52 cards, and there are 4 aces in the deck. The probability of drawing an ace first is given by the formula for probability: \( P(\text{Ace first}) = \frac{4}{52} \). Simplifying this fraction, we get \( \frac{1}{13} \).
02

Calculate Probability of Drawing a King Second

After the first card (an ace) is drawn, there are 51 cards left in the deck. Out of these remaining cards, there are 4 kings. Therefore, the probability of drawing a king second is \( P(\text{King second | Ace first}) = \frac{4}{51} \).
03

Calculate Combined Probability

Since we want to find the probability of both events happening in sequence, we multiply their individual probabilities. Thus, \( P(\text{Ace first and King second}) = \frac{1}{13} \times \frac{4}{51} = \frac{4}{663} \).
04

Construct Sample Space for Dice

Each die has 6 faces, so the sample space for two dice is all the ordered pairs \((x,y)\) where \(x\) and \(y\) can each be any integer from 1 to 6. Therefore, the sample space consists of 36 pairs: \((1,1), (1,2), \, \ldots, (6,6))\).
05

Calculate Probability of Same Values

There are 6 outcomes where the two dice show the same number: \((1,1), (2,2),..., (6,6)\). Thus, the probability that the two values are the same is \(\frac{6}{36} = \frac{1}{6}\).
06

Calculate Probability Values Differ by at Most One

Values differ by 0 or 1 in cases where the dice show same numbers and when one die is just one greater or less than the other. There are 11 such outcomes: \((1,1), (1,2), (2,1), (2,2),..., (6,6), (5,6), (6,5)\). Thus, the probability of the difference being at most 1 is \(\frac{11}{36}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Sample Space
In probability theory, a sample space represents the set of all possible outcomes of an experiment. Consider rolling two six-sided dice. Each die result ranges from 1 to 6. Thus, each roll has 6 possible outcomes. When rolling two dice, to find the sample space, form all possible pairs of results from the two dice:
  • Think of the dice as having two independent rolls.
  • Each die still has results from 1 through 6, making a total of 6 options per die.
Pairing these up, you create combinations like (1,1), (1,2) through (6,6). This leads to a total of 36 possible outcomes, forming what is called the sample space. Understanding the sample space is a crucial step to answer probability questions effectively.
Conditional Probability
Conditional probability explores the likelihood of an event occurring given that another event has already occurred. Let's say you draw a card from a deck. The chance of it being an ace is one question, but if it's drawn first and becomes an ace, the conditional probability then helps us calculate the chance of the next draw being a king.
  • Start with recognizing the first event: drawing the ace.
  • This initial event affects the sample space of the second event as there are now 51 cards left.
  • The occurrence of the first event influences our work in calculating the second event.
We consider this as a sequence: the first draw, identified as an ace reduces the deck's size, which makes now calculating the probability of drawing a king more relevant under this new condition. Thus, calculating these events together gives us the combined probability of both an ace and a king appearing in sequence.
Combinatorics
Combinatorics deals with counting and arrangement possibilities, especially crucial in probability when there are multiple ways events can occur. In the dice example, understanding how pairs are formed is what combinatorics helps us do, naturally linking it to finding probabilities.
  • Consider all possible outcomes or combinations鈥攍ike the pairs (x,y) when rolling dice.
  • Incorporating different constraints such as 'values differ by at most one.'
It's like knowing how to spread possible scenarios out on paper and capturing only the ones that match specific criteria. You utilize combinatorics techniques to discern patterns or sequences to aid in probability calculations and determine specific chances of events occurring.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The probability of issuing a drill of high brittleness (a reject) is \(0.02\). Drills are packed in boxes of 100 each. What is the probability that the number of defective drills is no greater than two?

The shelf life (in hours) of a certain perishable packaged food is a random variable with density function $$ f_{X}(x)=\left\\{\begin{array}{cl} 20000(x+100)^{-3} & (x>0) \\ 0 & \text { (otherwise) } \end{array}\right. $$ Find the probabilities that one of these packages will have a shelf life of (a) at least 200 hours; (b) at most 100 hours; (c) between 80 and 120 hours. The shelf life (in hours) of a certain perishable packaged food is a random variable with density function $$ f_{X}(x)=\left\\{\begin{array}{cl} 20000(x+100)^{-3} & (x>0) \\ 0 & \text { (otherwise) } \end{array}\right. $$ Find the probabilities that one of these packages will have a shelf life of (a) at least 200 hours; (b) at most 100 hours; (c) between 80 and 120 hours.

A manufacturer has agreed to dispatch small servomechanisms in cartons of 100 to a distributor. The distributor requires that \(90 \%\) of cartons contain at most one defective servomechanism. Assuming the Poisson approximation to the binomial distribution, write down an equation for the Poisson parameter \(\lambda\) such that the distributor's requirements are just satisfied. Solve by trial and error (approximate solution \(0.5\) ), and hence find the required proportion of manufactured servomechanisms that must be satisfactory.

The distribution of downtime \(T\) for breakdowns of a computer system is given by $$ f_{T}(t)=\left\\{\begin{array}{cl} a^{2} t \mathrm{e}^{-a t} & (t>0) \\ 0 & \text { (otherwise) } \end{array}\right. $$ where \(a\) is a positive constant. The cost of downtime derived from the disruption resulting from breakdowns rises exponentially with \(T\) : $$ \text { cost factor }=h(T)=\mathrm{e}^{b T} $$ Show that the expected cost factor for downtime is \([a /(a-b)]^{2}\), provided that \(a>b\)

Find the distribution of the sum of the numbers when a pair of dice is tossed.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Linear Momentum

Read Explanation

Torque and Rotational Motion

Read Explanation

Measurements

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.