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From a box containing 4 black balls and 2 green balls, 3 balls are drawn in succession, each ball being replaced in the box before the next draw is made. Find the probability distribution for the number of green balls.

Short Answer

Expert verified
The probability distribution for the number of green balls is: \(P(x=0) = \frac{8}{27}\), \(P(x=1) = \frac{12}{27}\), \(P(x=2) = \frac{6}{27}\), \(P(x=3) = \frac{1}{27}\)

Step by step solution

01

Understand Binomial Distribution

The binomial distribution is the discrete probability distribution of the number of successes in a sequence of n independent experiments. Here, the probability of drawing a green ball \(p = \frac{2}{6} = \frac{1}{3}\). Therefore we can use the formula for binomial distribution which is \(P(x=k) = C(n,k) * p^k * (1-p)^{n-k}\) where n is the total number of trials, k is the number of successful trials, C(n, k) is the number of combinations of n items taken k at a time, and p is the probability of success on a single trial.
02

Calculate for x=0,1,2,3

Calculate the binomial probabilities for x = 0, 1, 2 and 3. Where x is the random variable representing the number of green balls drawn. So,\n- \(P(x=0) = C(3,0) * (\frac{1}{3})^0 * (1- \frac{1}{3})^3 = \frac{8}{27}\)\n - \(P(x=1) = C(3,1) * (\frac{1}{3})^1 * (1- \frac{1}{3})^2 = \frac{12}{27}\)\n- \(P(x=2) = C(3,2) * (\frac{1}{3})^2 * (1- \frac{1}{3})^1 = \frac{6}{27}\)\n- \(P(x=3) = C(3,3) * (\frac{1}{3})^3 * (1- \frac{1}{3})^0 = \frac{1}{27}\)
03

Probability Distribution

Finally, the probability distribution for the number of green balls is as follows: \n- \(P(x=0) = \frac{8}{27}\)\n - \(P(x=1) = \frac{12}{27}\)\n - \(P(x=2) = \frac{6}{27}\)\n - \(P(x=3) = \frac{1}{27}\)

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Key Concepts

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

Binomial Theorem
Understanding the binomial theorem is essential in exploring the foundations of probability theory. At its core, the binomial theorem provides a method to expand algebraic expressions that are raised to a power, written in the form \( (a+b)^n \). Specifically, it states that this expression can be expanded into a sum involving terms of the form \( a^{n-k}b^k \) multiplied by coefficients that can be found in Pascal's triangle or calculated using combinations, symbolized as \( C(n,k) \).
For probability problems, the binomial theorem is the underpinning of binomial distribution, which predicts the number of successes in multiple trials of a binary process. In the provided exercise, the process of drawing a ball from a box is binary because the outcome is either drawing a green ball or not. Through the binomial theorem, you can foresee the various possible combinations when multiple draws are made, which is fundamental when determining the likelihood of these outcomes.
Probability Distribution
A probability distribution is a statistical function that describes all the possible outcomes of a random experiment and the likelihood of each event. When dealing with a finite set of discrete outcomes, the distribution is called a discrete probability distribution. In your example with the drawing of balls from a box, each draw is a discrete event with a finite number of outcomes.
The binomial probability distribution, a type of discrete distribution, is particularly relevant when you're dealing with a fixed number of independent trials of a binomial experiment. It's governed by two parameters: the number of trials (n) and the probability of success in a single trial (p). The key characteristic of a binomial distribution is that each trial is independent, and the probability of success remains constant throughout the trials, akin to 'each ball being replaced in the box before the next draw' in your exercise. This distribution forms a probability mass function, which shows the probability of achieving a certain number of successes (k) across the n trials.
Combinations
When calculating binomial probabilities, we often need to determine the number of ways to choose 'k' successes out of 'n' trials, regardless of order. This is where the concept of combinations comes into play. A combination is a selection of items from a larger pool where order doesn't matter. In contrast, a permutation would be used when the order does matter.
Mathematically, the number of combinations can be expressed using the combination formula \( C(n,k) = \frac{n!}{k!(n-k)!} \), where \( n! \) (n factorial) is the product of all positive integers up to n. In the context of our ball drawing exercise, \( C(n,k) \) would represent the different ways to draw 'k' green balls out of 'n' total draws. Understanding combinations is critical—they provide the coefficients for the binomial theorem and are fundamental in calculating probabilities for binomial distributions.
  • To visualize, think of building a sandwich from a variety of ingredients—it doesn't matter if you add lettuce before tomatoes or vice versa, it's the same sandwich.'
  • Similarly, in our exercise, drawing a green ball first, then black, and then green again is considered the same outcome as drawing green, green, then black when we're calculating the probability using combinations.

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Most popular questions from this chapter

Based on extensive testing, it is determined by the manufacturer of a washing machine that the time \(Y\) (in years) before a major repair is required is characterized by the probability density function $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{4} e^{-y / 4}, & y \geq 0, \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Critics would certainly consider the product a bargain if it is unlikely to require a major repair before the sixth year. Comment on this by determining \(P(Y>6)\) (b) What is the probability that a major repair occurs in the first year?

Let \(X\) denote the: number of heads and \(Y\) the number of heads minus the number of tails when 3 coins are tossed. Find the joint probability distribution of \(X\) and \(Y\).

An overseas shipment of 5 foreign automobiles contains 2 that have slight paint blemishes. If an agency receives 3 of these automobiles at random, list the elements of the sample space \(S\) using the letters \(B\) and \(N\) for blemished and nonblemished, respectively; then to each sample point assign a value \(x\) of the random variable \(X\) representing the number of automobiles purchased by the agency with paint blemishes.

Suppose it is known from large amounts of historical data that \(X\), the number of cars that arrive at a specific intersection during a 20 second time period, is characterized by the following discrete probability function $$ /(x)=e^{-6} \frac{b^{x}}{x !}, \quad x=0,1.2, \ldots $$ (a) Find the probability that in a specific 20 -second time period, more than 8 cars arrive at the intersection. (b) Find the probabilitythat only 2 cars arrive.

The time to failure in hours of an important piece of electronic equipment used in a manufactured DVD player las the density function $$ I(x)=\left\\{\begin{array}{ll} \frac{1}{2000} \exp (-a: / 2000), & x \geq 0, \\ 0, & x<0. \end{array}\right. $$ (a) Find \(F(x)\). (b) Determine the probability that the component (and thus the DVD player) lasts more than 1000 hours before the component needs to be replaced. (c) Determine the probability that the component fails before 2000 hours.

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