/*! 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 33 Express all probabilities as fra... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Express all probabilities as fractions. In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. A winning "blackjack" hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points. The two cards can be in any order. Find the probability of being dealt a blackjack hand. What approximate percentage of hands are winning blackjack hands?

Short Answer

Expert verified
The probability of being dealt a blackjack hand is \(\frac{32}{663}\), which is approximately 4.8\text{%}.

Step by step solution

01

Identify Total Possible Hands

First, calculate the total number of possible 2-card combinations from a deck of 52 cards. Use the combination formula \(\binom{n}{k}\) to determine the total number of combinations: \(\binom{52}{2}\).
02

Calculate Total Possible Hands

Use the combination formula to find the number of ways to choose 2 cards out of 52: \(\binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 1326\). There are 1326 possible 2-card hands.
03

Identify Winning Blackjack Hands

A winning blackjack hand consists of 1 of the 4 aces and 1 of the 16 cards worth 10 points (10, Jack, Queen, King). The number of such combinations is obtained by multiplying: 4 aces \(\times\) 16 ten-point cards.
04

Calculate Winning Blackjack Hands

Number of winning combinations = 4 (Aces) \(\times\) 16 (ten-point cards) = 64. There are 64 possible winning combinations.
05

Compute Probability of Winning Blackjack Hand

The probability of being dealt a blackjack hand is the ratio of the number of winning hands to the total number of possible hands: \(\frac{64}{1326}\). Simplify the fraction if possible.
06

Simplify the Fraction

Simplify \(\frac{64}{1326}\) by finding the greatest common divisor of 64 and 1326, which is 2. Divide both the numerator and the denominator by 2 to get \(\frac{32}{663}\). Thus, the probability is \(\frac{32}{663}\).
07

Find Approximate Percentage

Convert the fraction \(\frac{32}{663}\) to a percentage by multiplying it by 100. \(\frac{32}{663} \times 100 \approx 4.8\text{%.}\).

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 91Ó°ÊÓ!

Key Concepts

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

combination formula
To find probabilities in games like blackjack, it's essential to understand the concept of combinations. The combination formula helps us determine the number of ways to choose a subset of items from a larger set. For a blackjack hand, we'll use the combination formula \( \binom{n}{k} \), where \( n \) is the total number of items, and \( k \) is the number of items to choose. 
For example, to find the number of ways to choose 2 cards from a 52-card deck, we use \( \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 1326 \). This tells us there are 1326 possible 2-card combinations in a standard deck.
It's crucial to note that the combination formula is different from permutations. Combinations do not consider the order of selection, while permutations do.
blackjack probabilities
In blackjack, the goal is to achieve a hand value as close to 21 as possible, without exceeding that value. A blackjack hand is specifically an Ace paired with a 10-point card (10, Jack, Queen, King).
To determine the probability of being dealt a blackjack hand, we start by identifying all winning combinations. We have 4 aces and 16 ten-point cards. There are \( 4 \times 16 = 64 \) possible blackjack hands.
Next, we've already calculated that there are 1326 possible 2-card combinations in the whole deck. Hence, the probability of getting a blackjack hand is \( \frac{64}{1326} \). After simplifying, this fraction becomes \( \frac{32}{663} \).
Blackjack probabilities can also be expressed as percentages for easier understanding. By converting this fraction to a percentage, we get approximately 4.8%.
fraction simplification
Simplifying fractions is essential in probability to make the results clearer and more concise. The original fraction we obtained was \( \frac{64}{1326} \), which can be a bit unwieldy. By finding the greatest common divisor (GCD) of the numerator and the denominator, we can simplify the fraction.
For \( \frac{64}{1326} \), the GCD is 2. So, we divide both the numerator and the denominator by 2, resulting in \( \frac{32}{663} \). This simplified fraction is much easier to work with and interpret.
Fraction simplification not only helps in easier calculation but also aids in better comparison and understanding of probabilities.
percentage calculation
After finding the simplified fraction for the probability, it's often useful to convert that fraction into a percentage. Percentages are more intuitive and easier for most people to understand.
To convert a fraction to a percentage, you multiply it by 100. For the simplified fraction \( \frac{32}{663} \), we multiply by 100: \( \frac{32}{663} \times 100 \approx 4.8\% \).
This tells us that approximately 4.8% of all possible 2-card hands dealt in blackjack will be winning blackjack hands. Understanding percentage calculation is crucial for interpreting statistical probabilities in a clear and concise manner.

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

Find the probability and answer the questions. Men have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective \(X\) chromosome that occurs without a paired \(X\) chromosome that is not defective. In the following, represent a defective \(X\) chromosome with lowercase \(x\), so a child with the \(x Y\) or Yx pair of chromosomes will have the disease and a child with \(X X\) or \(X Y\) or \(\mathrm{YX}\) or \(\mathrm{xX}\) or \(\mathrm{Xx}\) will not have the disease. Each parent contributes one of the chromosomes to the child. a. If a father has the defective \(x\) chromosome and the mother has good XX chromosomes, what is the probability that a son will inherit the disease? b. If a father has the defective \(x\) chromosome and the mother has good XX chromosomes, what is the probability that a daughter will inherit the disease? C. If a mother has one defective \(x\) chromosome and one good \(X\) chromosome and the father has good XY chromosomes, what is the probability that a son will inherit the disease? d. If a mother has one defective \(x\) chromosome and one good \(X\) chromosome and the father has good XY chromosomes, what is the probability that a daughter will inherit the disease?

When randomly selecting an adult, A denotes the event of selecting someone with blue eyes. What do \(P(A)\) and \(P(\bar{A})\) represent?

Find the probability. Subjects for the next presidential election poll are contacted using telephone numbers in which the last four digits are randomly selected (with replacement). Find the probability that for one such phone number, the last four digits include at least one \(0 .\)

High Fives a. Five "mathletes" celebrate after solving a particularly challenging problem during competition. If each mathlete high fives each other mathlete exactly once, what is the total number of high fives? b. If \(n\) mathletes shake hands with each other exactly once, what is the total number of handshakes? c. How many different ways can five mathletes be seated at a round table? (Assume that if everyone moves to the right, the seating arrangement is the same.) d. How many different ways can \(n\) mathletes be seated at a round table?

Express all probabilities as fractions. Your professor has just collected eight different statistics exams. If these exams are graded in random order, what is the probability that they are graded in alphabetical order of the students who took the exam?

See all solutions

Recommended explanations on Math Textbooks

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.