Chapter 15: Problem 9

(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 \(\sigma\)

Short Answer

Expert verified

Sample space: 0, 1, 2. Probability of at least one 3: 3/8. P(Box1 | at least one 3): 1/3. \bar{x} = 5/24, \sigma ≈ 0.44.

Step by step solution

Sample Space Set Up

Identify the sample space and outcomes for each box. For Box 1 (one die), possible outcomes are 0 or 1. For Box 2 (two dice), possible outcomes are 0, 1, or 2.

Calculate Individual Probabilities

Each die has a probability of \(1/6\) for landing on 3. For Box 1: \(P(x=0) = 5/6\) and \(P(x=1) = 1/6\). For Box 2: \(P(x=0) = (5/6)^2 = 25/36\), \(P(x=1) = 2* (1/6) * (5/6) = 10/36\), and \(P(x=2) = 1/36\).

Combine Probabilities with Box Selection

The probability of picking either box is \(1/2\). Combine this with outcomes: \(P(x=0) = 1/2 * 5/6 + 1/2 * 25/36 = 55/72\), \(P(x=1) = 1/2 * 1/6 + 1/2 * 10/36 = 13/36\), and \(P(x=2) = 1/2 * 1/36 = 1/72\).

Probability of at Least One 3

Sum the probabilities of outcomes with at least one 3: \(P(x \geq 1) = P(x=1) + P(x=2) = 13/36 + 1/72 = 27/72 = 3/8\).

Conditional Probability of Picking Box 1

Use Bayes' Theorem: \(P(Box1|x \geq 1) = (P(x \geq 1|Box1) * P(Box1))/(P(x \geq 1))\) where \(P(x \geq 1|Box1) = 1/6\) and \(P(x \geq 1) = 3/8\). Calculate: \(P(Box1|x \geq 1) = ((1/6 * 1/2)/(3/8)) = 1/3\).

Calculate Expected Value (\(\bar{x}\)) and Variance (\(\sigma\))

Use the formula for expected value: \bar{x} = \( \sum (x * P(x)) = 0 * 55/72 + 1 * 13/36 + 2 * 1/72 = 15/72 = 5/24\). For variance \(\sigma^2, \sigma^2 = \sum (x-\bar{x})^2 * P(x) \: \sigma^2 = (0-5/24)^2 * 55/72 + (1-5/24)^2 * 13/36 + (2-5/24)^2 * 1/72 \), which simplifies to approximately 0.19. Thus, \sigma ≈ 0.44\.

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Key Concepts

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

Sample Space

In probability theory, a sample space is the set of all possible outcomes of an experiment. For the given exercise:

Box 1 contains one die, so the possible outcomes for the number of threes showing (\(x\)) are 0 or 1.
Box 2 contains two dice, so the possible outcomes for \(x\) are 0, 1, or 2.

To summarize, the sample space can be written as {0, 1, 2}.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. In the exercise:

We need to determine the probability of picking Box 1 given that at least one 3 has shown up.

This is denoted as \(P(Box1 \,| x ≥ 1)\). Using Bayes' Theorem, `\(P(Box1 \,| x ≥ 1 ) = \frac{P(x ≥ 1 \,| Box1) * P(Box1)}{P(x ≥ 1)}\). We plug in our calculated values: \(P(x ≥ 1|Box1) = \frac{1}{6}\), \(P(Box1) = \frac{1}{2}\), and \(P(x ≥ 1) = \frac{3}{8}\). This results in \(P(Box1|x≥1) = \frac{1/12}{3/8} = \frac{1}{3}\).

Expected Value

The expected value (average or mean) is a measure of the center of a probability distribution. For our exercise, we use the formula:
\[ \bar{x} = \sum (x * P(x)) \] where \(P(x)\) are the probabilities calculated for each outcome. Thus: \[ \bar{x} = 0 * \frac{55}{72} + 1 * \frac{13}{36} + 2 * \frac{1}{72} = \frac{5}{24} \]

Variance

Variance measures how much the values of a data set differ from the expected value. It's calculated by: \[ \sigma^2 = \sum (x - \bar{x})^2 * P(x) \] Using our values: \[ \sigma^2 = (0-\frac{5}{24})^2 * \frac{55}{72} + (1-\frac{5}{24})^2 * \frac{13}{36} + (2-\frac{5}{24})^2 * \frac{1}{72} \] This approximates to 0.19, and hence, the standard deviation \(\sigma ≈ 0.44\).

Bayes' Theorem

Bayes' Theorem relates the conditional probability of two events. For our scenario, it helps find the probability of selecting the first box given that at least one 3 is shown. The theorem is stated as: \[P(A|B) = \frac{P(B|A) * P(A)}{P(B)} \] With our inputs \(P(A)\) being the probability of picking the first box, \(P(B|A)\) being the probability of at least one 3 given the first box, and \(P(B)\) being the overall probability of at least one 3. We obtain: \[P(Box1 | x \geq 1) = \frac{(1/6 * 1/2)}{3/8} = 1/3. \]