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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

Let \(X\) denote the diameter of an armored electric cable and \(Y\) denote the diameter of the ceramic mold that makes the cable. Both \(X\) and \(Y\) are scaled so that they range between 0 and \(1 .\) Suppose that \(X\) and \(Y\) have the joint density $$ f(x, y)=\left\\{\begin{array}{ll} \frac{1}{y}, & 0< x < y<1; \\ 10, & \text { elsewhere. } \end{array}\right. $$ Find \(P(X+Y>1 / 2)\).

Magnetron tulses are produced from an automated assembly line. A sampling plan is used periodically to assess quality on the lengths of the tubes. This measurement is subject to uncertainty. It is thought that the probability that a random tube meets length specification is \(0.99 .\) A sampling plan is used in which the lengths of 5 random tubes are measured. (a) Show that the probability function of \(Y\), the number out of 5 that meet length specification, is given by the following discrete probability function $$ \begin{aligned} f(y) &=\frac{5 !}{y !(5-y) !}(0.99)^{y}(0.01)^{5-y.} \\ \text { for } y=0.1,2,3,4,5. \end{aligned} $$ (b) Suppose random selections are made off the line and 3 are outside specifications. Use \(f(y)\) above either to support or refute the conjecture that the probability is 0.99 that a single tube meets specifications.

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 waiting time, in hours, between successive speeders spotted by a radar unit is a continuous random variable with cumulative distribution function $$ F(x)=\left\\{\begin{array}{ll} 0_{+} & x<0, \\ 1-e^{-k x}, & x \geq 0. \end{array}\right. $$ Find the probability of waiting less than 12 minutes between successive speeders (a) using the cumulative distribution function of \(X\); (b) using the probability density function of \(X\).

An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by $$ /(x)=\left\\{\begin{array}{ll} 3 x^{-4}, & x>1, \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Verify that this is a valid density function. (b) Evaluate \(F(x)\). (c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?

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