In a box there are 2 white, 3 black, and 4 red balls. If a ball is drawn at random, what is the probability that it is black? That it is not red?

Short Answer

Expert verified
The probability of drawing a black ball is \( \frac{1}{3} \). The probability of drawing a ball that is not red is \( \frac{5}{9} \).

Step by step solution

01

Calculate Total Number of Balls

Count the total number of balls in the box. There are 2 white, 3 black, and 4 red balls. Thus, the total number is calculated as follows: \[ \text{Total number of balls} = 2 + 3 + 4 = 9 \]
02

Calculate the Probability of Drawing a Black Ball

The probability of drawing a black ball is the ratio of the number of black balls to the total number of balls. The number of black balls is 3. Thus, the probability is calculated as follows: \[ P(\text{Black}) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{3}{9} = \frac{1}{3} \]
03

Calculate the Probability of Drawing a Ball That Is Not Red

The number of balls that are not red is the total number of balls minus the number of red balls. There are 4 red balls. Thus, the number of non-red balls is: \[ 9 - 4 = 5 \] The probability of drawing a ball that is not red is the ratio of non-red balls to the total number of balls. Hence, the probability is: \[ P(\text{Not Red}) = \frac{\text{Number of non-red balls}}{\text{Total number of balls}} = \frac{5}{9} \]

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Key Concepts

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

probability calculation
Probability calculation is a fundamental aspect of probability theory that helps us understand the likelihood of events occurring. In this exercise, we calculated the probability of drawing a black ball and a non-red ball from a box containing white, black, and red balls.
To do this, follow these steps:
1. Count the total number of elements (in this case, balls) in the set.
2. Identify the subset of elements representing the event of interest (e.g., black balls or non-red balls).
3. Calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes.

For example, the probability of drawing a black ball was calculated as follows: \( P(\text{Black}) = \frac{3}{9} = \frac{1}{3} \)
Similarly, the probability of drawing a ball that is not red was calculated as: \( P(\text{Not Red}) = \frac{5}{9} \)

Always remember to simplify fractions whenever possible for a clearer and more concise result.
combinatorics
Combinatorics is the branch of mathematics dealing with combinations, permutations, and counting. It is essential in probability theory to help determine the number of possible outcomes and favorable outcomes.

Here are a few key concepts in combinatorics:
  • **Combination**: Selection of items where order does not matter.
  • **Permutation**: Arrangement of items where order does matter.
  • **Counting Principle**: Using multiplication to find the number of combinations.
In this exercise, we are not explicitly using combinatorics. However, understanding how to count the total number of balls (9 in this case) and the number of black balls (3) or non-red balls (5) relies on basic counting principles.

Remember, always ensure accurate counting of the entire set and subsets representing the events of interest to accurately apply probability calculations.
basic statistics
Basic statistics involves the collection, analysis, interpretation, and presentation of data. In the context of this exercise, we used simple statistical measures to calculate probabilities.

Understanding the idea of 'proportion' is crucial. A proportion represents the fraction of a subset within the entire set.
  • Calculation: Proportion = \(\frac{Number \ of \ subset \ elements}{Total \ number \ of \ elements} \)
  • Presentation: Simplify the fraction for clarity.
For example, the proportion of black balls is \(\frac{3}{9}\), which simplifies to \(\frac{1}{3}\). This expresses that one third of the total balls are black.

Similarly, the probability of drawing a ball that is not red is calculated by determining the number of non-red balls and then forming a proportion: \(\frac{5}{9}\).

Therefore, understanding and using basic statistical methods help in deriving meaningful conclusions from data, especially when calculating probabilities.
random events
Random events are those that occur without a specific pattern or predictability. In probability theory, we often deal with random events to measure their likelihood.

Key points about random events:
  • **Independent Events**: Events where the occurrence of one does not affect the other.
  • **Dependent Events**: Events where the occurrence of one affects the other.
  • **Random Variables**: Numerical outcomes of random events (e.g., drawing a black ball).
In this exercise, drawing a ball from the box is a random event since each ball has an equal chance of being selected.

The probability of drawing a black ball was calculated by identifying the favorable outcomes (black balls) and comparing them to the total number of possible outcomes (all balls in the box). This simple example demonstrates the fundamental nature of random events and the importance of probability theory in understanding their behavior.

Random events are essential to many areas of study and practical applications, from simple games to complex financial models.

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Most popular questions from this chapter

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) One box contains one die and another box contains two dice. You select a box at random and take out and toss whatever is in it (that is, toss both dice if you have picked box 2 ). Let \(x=\) number of 3 's showing. Set up the sample space and associated probabilities for \(x\). (b) What is the probability of at least one 3 ? (c) If at least one 3 turns up, what is the probability that you picked the first box? (d) Find \(\bar{x}\) and \(\sigma\).

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?

Would you pay \(\$ 10\) per throw of two dice if you were to reccive a number of dollars equal to the product of the numbers on the dice? Himt: What is your expectation? If it is more than \(\$ 10\), then the game would be favorable for you.

(a) A candy vending machine is out of order. The probability that you get a candy bar (with or without the return of your quarter) is \(\frac{1}{2}\), the probability that you get your quarter back (with or without candy) is \(\frac{1}{3}\), and the probability that you get both the candy and your money back is \(\frac{1}{12}\). What is the probability that you get nothing at all? Sugrestion: Sketch a geometric diagram similar to Figure \(3.1\), indicate regions representing the various possibilities and their probabilities; then set up a four-point samplc space and the associated probabilities of the points. (b) Suppose you. put another quarter into the candy vending machine of part (a). Set up the 16-point sample space corresponding to the possible results of your two attempts to bu? a candy bar, and find the probability that you get two candy bars (and no money back): that you get no candy and lose both quarters; that you just get your money back both times.

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