/*! 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 2 You are dealt one card from a 52... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 2

You are dealt one card from a 52-card deck. Find the probability that you are not dealt a \(3 .\)

Short Answer

Expert verified
The probability of not drawing a 3 from a 52-card deck is \(\frac{48}{52} = 0.923\).

Step by step solution

01

Identify Total Number of Outcomes

Understand that a standard deck of cards contains 52 cards. This means that there are 52 possible outcomes when a card is drawn at random from the deck.
02

Identify Desired Outcomes

Recognize that a deck of 52 cards is composed of 4 cards each from numbers 2-10, and face cards (Jack, Queen, King, Ace). This totals to 4 cards each for 13 different characters or numbers, including 3. So, there are 4 cards with the number 3, meaning there are \(52 - 4 = 48\) cards that are not 3.
03

Calculate the Probability

The probability is calculated by dividing the total number of desired outcomes (cards that are not 3) by the total number of possible outcomes (total cards in the deck). Therefore, the probability \(P\) of drawing a card that is not a 3 is given by the formula: \(P = \frac{number\: of\: desired\: outcomes}{total\: number\: of\: outcomes} = \frac{48}{52}\).

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!

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

Involve computing expected values in games of chance. For many years, organized crime ran a numbers game that is now run legally by many state governments. The player selects a three-digit number from 000 to 999 . There are 1000 such numbers. A bet of \(\$ 1\) is placed on a number, say number 115. If the number is selected, the player wins \(\$ 500\). If any other number is selected, the player wins nothing. Find the expected value for this game and describe what this means.

An ice chest contains six cans of apple juice, eight cans of grape juice, four cans of orange juice, and two cans of mango juice. Suppose that you reach into the container and randomly select three cans in succession. Find the probability of selecting no grape juice.

Make Sense? Determine whether each statement makes sense or does not make sense, and explain your reasoning. The Fundamental Counting Principle can be used to determine the number of ways of arranging the numbers \(1,2,3,4,5, \ldots, 98,99,100\)

We return to our box of chocolates. There are 30 chocolates in the box, all identically shaped. Five are filled with coconut, 10 with caramel, and 15 are solid chocolate. You randomly select one piece, eat it, and then select a second piece. Find the probability of selecting a coconut-filled chocolate followed by a solid chocolate.

In the original plan for area codes in 1945 , the first digit could be any number from 2 through 9 , the second digit was either 0 or 1 , and the third digit could be any number except 0 . With this plan, how many different area codes are possible?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Probability and Statistics

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.