Chapter 16: 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}. P(at least 1 three) = \(\frac{17}{72}\). If at least one 3: \(\frac{3}{17}\). \(\bar{x}=\frac{7}{18}\), \(\sigma\approx \frac{11}{36}\).

Step by step solution

Identify the Sample Space

First, determine all possible outcomes when selecting either box 1 or box 2. For box 1 (one die), the possible outcomes are: {1, 2, 3, 4, 5, 6}. For box 2 (two dice), the total possible outcomes would be combinations of values from two dice, leading to arrays such as {1,1}, {1,2}, ..., {6,6}.

Define Random Variable x

Let the random variable \(x\) denote the number of 3's showing on the die/dice. Evaluate \(x\) for both box 1 and box 2. For box 1, \(x = 0\) if any number other than 3 shows up, and \(x = 1\) if 3 shows up. For box 2, outcomes could be {x=0, x=1, x=2}, representing no 3's, one 3, or two 3's respectively.

Calculate Probabilities for Each x

Calculate the probability of each \(x\) value. For box 1 (one die), \(P(x=0)=\frac{5}{6}\), \(P(x=1)=\frac{1}{6}\). For box 2 (two dice), use the binomial probability formula to find: \(P(x=0)=\frac{25}{36}\), \(P(x=1)=\frac{10}{36}\), \(P(x=2)=\frac{1}{36}\). Given that choosing either box is equally likely, weigh these probabilities by 0.5.

Compute Overall Probabilities

Combine the weighted probabilities of each \(x\) value for both boxes. Use: \[P(x=k) = 0.5 \times P_{box1}(x=k) + 0.5 \times P_{box2}(x=k)\]. Calculate: \(P(x=0) = 0.5 \times \frac{5}{6} + 0.5 \times \frac{25}{36} = \frac{55}{72}\), \(P(x=1) = 0.5 \times \frac{1}{6} + 0.5 \times \frac{10}{36} = \frac{3}{12} = \frac{18}{72}\), and \(P(x=2) = 0 + 0.5 \times \frac{1}{36} = \frac{1}{72}\).

Find Probability of At Least One 3 (Part b)

To find the probability of at least one 3, calculate \(P(x eq 0) = 1 - P(x=0)\). Use previous results, \(P(x=0) = \frac{55}{72}\), so find \(P(x eq 0) = 1 - \frac{55}{72} = \frac{17}{72}\).

Conditional Probability for Part c

Given at least one 3, apply Bayes' Theorem to find the probability you picked box 1, \(P(Box 1|x eq 0) = \frac{P(x eq 0|Box 1) \times P(Box 1)}{P(x eq 0)}\). Simplify values: \(P(x eq 0|Box 1) = \frac{1}{6}\), \(P(Box 1) = 0.5\), and from Step 5, \(P(x eq 0) = \frac{17}{72}\). This gives \[P(Box 1|x eq 0) = \frac{\frac{1}{6} \times 0.5}{\frac{17}{72}} = \frac{3}{17}\].

Calculate Mean and Standard Deviation for Part d

To find the expected value and standard deviation of \(x\): - Expected value, \(\bar{x}\): \[\bar{x} = \frac{5}{36} + \frac{7}{36} + \frac{2}{36} = \frac{14}{36} = \frac{7}{18}\]. - Variance: \(\sigma^2 = \sum (x_i - \bar{x})^2 P(x_i)\), calculate each term and sum. - Standard deviation: \(\sigma = \sqrt{\sigma^2}\).

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Key Concepts

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

Sample Space

A sample space is simply the set of all possible outcomes of an experiment. For instance, if you flip a coin, your sample space is {Heads, Tails}. In our exercise, we are dealing with two boxes. The first box contains one die, and the second box contains two dice.
When you roll one die from the first box, the possible outcomes are {1, 2, 3, 4, 5, 6}.
When you roll two dice from the second box, each die can show any number between 1 and 6, making the total number of outcomes 36. The sample space in this case is all combinations of these pairs, like {1,1}, {1,2}, ..., {6,6}.
This helps us to list out all the possible results before calculating the probabilities involved in each.

Random Variable

A random variable is a variable that can take on different values based on the outcomes of a random experiment. In our example, the random variable, often denoted as \(x\), represents the number of 3's showing when tossing the dice.
For the first box (one die), \(x\) can either be 0 (if the outcome is anything other than 3) or 1 (if the die shows 3).
For the second box (two dice), \(x\) can be 0 (no 3's show), 1 (one of the dice shows 3), or 2 (both dice show 3).
By assigning these values, the random variable helps us quantify the results of the experiment in a clear and structured way, making it easier to solve problems with probability theory.

Expected Value

The expected value, also known as the mean, is a measure of the central tendency of a random variable. In simple terms, it tells us what's the 'average' or 'most likely' outcome if we were to repeat the experiment many times. The expected value of a random variable \(x\) is calculated using: \(\bar{x} = \text{sum of all} \ (x_i \times P(x_i)) \), where \(x_i\) are the possible values and \(P(x_i)\) are their probabilities.
In our exercise, to find the expected number of 3's, \(x\), we need to combine probabilities from both boxes. Using the values we calculated earlier, the expected value \(\bar{x}\) = \frac{7}{18}\. \ . This gives us a comprehensive understanding of what to expect on average when repeating this experiment multiple times.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how spread out the outcomes of a random variable are from the mean (expected value). A higher standard deviation indicates that the values are more spread out. To find the standard deviation, we first need to calculate the variance, which is the average of the squared differences from the mean.
For this, use the formula: \(\text{Variance} = \text{sum of all} \ ((x_i - \bar{x})^2 \times P(x_i))\) . Then, the standard deviation (\