Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Give the probability, as a percentage, of picking the indicated card from a deck. 2 or 3

Short Answer

Expert verified
Answer: The probability is approximately 15.38%.

Step by step solution

01

Calculate the number of possible outcomes and the number of successful outcomes

In a standard deck of 52 playing cards, there are 4 of each card (4 suits). So, there are 4 twos and 4 threes. Number of successful outcomes = 4 + 4 = 8 (either a 2 or a 3) Total possible outcomes = 52 (cards in the deck)
02

Calculate the probability of picking a 2 or a 3

To calculate the probability of an event, we can use the formula: Probability = (Number of successful outcomes) / (Total number of possible outcomes) In our case, the probability of picking a 2 or a 3 is: Probability = 8 / 52
03

Convert the probability to a percentage

To convert the probability to a percentage, we can multiply the probability by 100. Percentage = Probability * 100 Percentage = (8 / 52) * 100
04

Simplify and find the final answer

Now we can simplify the fraction and find the percentage: Percentage = (8 / 52) * 100 = (2 / 13) * 100 ≈ 15.38% Thus, the probability of picking a 2 or a 3 from a deck of 52 cards is approximately 15.38%.

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 Vaia!

Key Concepts

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

Percentage Calculation
When we talk about percentages, we're describing a portion out of 100. A percentage tells us how many parts of a whole are taken if the whole is divided into 100 equal parts. In probability, converting a decimal to a percentage can make it easier to understand. Here's how you do it:
  • First, calculate the probability as a fraction.
  • Multiply that fraction by 100 to convert it into a percentage.
For example, if the probability of picking a 2 or a 3 from a deck is \( \frac{8}{52} \), to find the percentage, you compute \( \left(\frac{8}{52}\right) \times 100 \). Simplifying \( \frac{8}{52} \) to \( \frac{2}{13} \), the calculation becomes \( \left(\frac{2}{13}\right) \times 100 \). Your final result is approximately 15.38%, showing a clear and understandable way to express the likelihood of the event.
Playing Cards
A standard deck of playing cards consists of 52 cards and is divided into four suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards: numbered from 2 to 10, a Jack, a Queen, a King, and an Ace. To discuss card probability, remember:
  • Each card drawn from the deck represents a unique outcome.
  • The deck is typically in a well-shuffled state, meaning each card has an equal chance of being picked.
In this context, knowing these details helps solve problems like finding the probability of drawing either a 2 or a 3 from the deck. Since there are 4 twos and 4 threes, the combination of these successful outcomes (8 cards) is crucial when calculating probabilities in card games.
Successful Outcomes
In probability, a successful outcome refers to the event of interest happening based on our criteria. To solve problems, identify the number of successful outcomes and compare them to the total number of possible outcomes.Here’s how you do it:
  • Determine what outcomes qualify as a success - in this example, picking a 2 or a 3.
  • Count these outcomes. For a standard deck, there are 4 twos and 4 threes, totaling 8 successful outcomes.
Then, use the formula to find the probability: Probability = \( \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}} \). This direct comparison simplifies understanding the likelihood of specific card events, enhancing your approach to these probability questions.

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 mean of \(1,2,3, \ldots, n,\) if \(n\) is (a) 3 (b) 4 (c) 5 (d) 6 (e) \(2 \mathrm{k}\) (f) \(2 \mathrm{k}+1\)

Suppose two samples of values are taken from a population $$ a: 8,2,4,7,5 \text { and } b: 6,9,10,5,7,8,2,5 $$ (a) Find \(\bar{a}\) and \(\bar{b}\). (b) Find the mean of the sample you get by combining the two samples. (c) Is the mean of the combined sample equal to the mean of the two values \(\bar{a}\) and \(\bar{b}\) ? (d) Explain why the means in (c) are the same or different.

Find the mean of the data set. $$ 2,4,6,8 $$

Face recognition systems pick faces out of crowds at airports to see if any matches occur with law enforcement databases. Performance of the systems can be affected by lighting, gender and age of the target, and age of the database. In Problems \(24-27,\) the tables give identification rates for faces under various conditions. Decide whether the rate of recognition is independent of the given factor. Table 17.22 compares face recognition under different lighting conditions.$$ \begin{array}{c|c|c} \hline \text { Lighting } & \text { Did recognize } & \text { Did not recognize } \\ \hline \text { Indoors } & 900 & 100 \\ \hline \text { Outdoors } & 300 & 300 \\ \hline \end{array} $$

A sample of \(n_{1}\) frogs has a total weight of \(W_{1}\) grams. A second sample of \(n_{2}\) frogs has a total weight of \(W_{2}\) grams. (a) What is the mean weight (in grams per frog) of each sample? (b) What is the mean weight of all the frogs in both samples? Does it equal the average of the mean weights of the samples taken separately?

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.

Sign-up for free