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There are 3 red and 2 white balls in one box and 4 red and 5 white in the second box. You select a box at random and from it pick a ball at random. If the ball is red, what is the probability that it came from the second box?

Short Answer

Expert verified
The probability is approximately 0.460.

Step by step solution

01

- Understand the problem

Need to find the probability that a red ball came from the second box given that it was chosen at random.
02

- Define Events

Let A be the event that Box 1 is chosen, and B be the event that Box 2 is chosen. Let R be the event that a red ball is chosen.
03

- Determine Probabilities for Choosing Each Box

Since a box is chosen at random, the probability of choosing either box is equal: \[ P(A) = P(B) = \frac{1}{2} \]
04

- Determine Probabilities of Red Ball from Each Box

The probability of picking a red ball given that Box 1 was chosen is: \[ P(R|A) = \frac{3}{3+2} = \frac{3}{5} \] The probability of picking a red ball given that Box 2 was chosen is: \[ P(R|B) = \frac{4}{4+5} = \frac{4}{9} \]
05

- Use Bayes' Theorem

Using Bayes' Theorem, the probability that the red ball came from the second box is given by: \[ P(B|R) = \frac{P(R|B)P(B)}{P(R)} \]
06

- Find Total Probability of Red Ball

The total probability of drawing a red ball, P(R), is: \[ P(R) = P(R|A)P(A) + P(R|B)P(B) = \frac{3}{5} \times \frac{1}{2} + \frac{4}{9} \times \frac{1}{2} = \frac{3}{10} + \frac{4}{18} \approx 0.483 \]
07

- Calculate Final Probability

Substitute back into Bayes' Theorem: \[ P(B|R) = \frac{\frac{4}{9} \times \frac{1}{2}}{0.483} \approx 0.460 \]

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Key Concepts

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

Bayes' Theorem
Bayes' Theorem is a powerful tool in probability theory that allows you to update the probability estimate for an event based on new evidence. Named after Thomas Bayes, it helps you find the probability of an event given that another event has occurred.
To understand Bayes' Theorem, it's essential to know the formula:
\[P(A|B) = \frac{P(B|A)P(A)}{P(B)}\]
Here's what each term means:
  • \(P(A|B)\) is the probability of event A occurring given that event B has occurred.
  • \(P(B|A)\) is the probability of event B occurring given that event A has occurred.
  • \(P(A)\) and \(P(B)\) are the probabilities of events A and B occurring independently of each other.
In our exercise, we want to determine the probability that a red ball came from the second box (event B) given that a red ball was picked (event R). Bayes' Theorem helps us update our initial beliefs based on this new information.
Probability Theory
Probability Theory is a branch of mathematics that deals with the analysis of random events. It provides a mathematical foundation to quantify and understand the likelihood of different outcomes.
In our exercise, we deal with several basic concepts of probability theory:
  • Probability of an event: This measures how likely an event is to occur. For example, the probability of choosing either box at random is \(\frac{1}{2}\).
  • Conditional Probability: This measures the probability of an event occurring given that another event has already occurred. For instance, the probability of picking a red ball from the second box given that you chose the second box is \(\frac{4}{9}\).
Essential formulas used in this exercise include:
  • Probability of picking a red ball from the first box (\(P(R|A)\)) and the second box (\(P(R|B)\)).
  • Total probability of drawing a red ball (\(P(R)\)), calculated by:
    \[P(R) = P(R|A)P(A) + P(R|B)P(B)\]
This structured approach helps simplify complex problems and apply logical methods to solve them.
Random Selection
Random Selection is a process where each item in a set has an equal chance of being chosen. This concept is central to many problems in probability theory, including our exercise.
In the context of our problem, randomness appears in two stages:
  • Choosing one of the two boxes (first or second) where the probability of each box being chosen is \(\frac{1}{2}\).
  • Picking a ball from the chosen box, where each ball has an equal chance of being selected.
This randomness ensures that the problem is fair and unbiased, allowing us to apply probability theory accurately.
To make informed decisions, it's crucial to understand the role of randomness. For the exercise, knowing that each box has the same chance of being chosen simplifies parts of the calculation. Random selection helps break down seemingly complicated problems into understandable steps, paving the way for applying conditional probability and Bayes' Theorem.

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Most popular questions from this chapter

Let \(m_{1}, m_{2}, \cdots, m_{n}\) be a set of measurements, and define the values of \(x_{i}\) by \(x_{1}=\) \(m_{1}-a, x_{2}=m_{2}-a, \cdots, x_{n}=m_{n}-a,\) where \(a\) is some number (as yet unspecified, but the same for all \(x_{i}\) ). Show that in order to minimize \(\sum_{i=1}^{n} x_{i}^{2},\) we should choose \(a=(1 / n) \sum_{i=1}^{n} m_{i} .\) Hint: Differentiate \(\sum_{i=1}^{n} x_{i}^{2}\) with respect to \(a .\) You have shown that the arithmetic mean is the "best" average in the least squares sense, that is, that if the sum of the squares of the deviations of the measurements from their "average" is a minimum, the "average" is the arithmetic mean (rather than, say, the median or mode).

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

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). Suppose that Martian dice are 4-sided (tetrahedra) with points labeled 1 to 4. When a pair of these dice is tossed, let \(x\) be the product of the two numbers at the tops of the dice if the product is odd; otherwise \(x=0\).

Let \(\mu\) be the average of the random variable \(x\). Then the quantities \(\left(x_{i}-\mu\right)\) are the deviations of \(x\) from its average. Show that the average of these deviations is zero. Hint: Remember that the sum of all the \(p_{i}\) must equal 1.

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). A weighted coin with probability \(p\) of coming down heads is tossed three times; \(x=\) number of heads minus number of tails.

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