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Chapter 7: Problem 22

Another die is weighted in such a way that each of 1 and 2 is three times as likely to come up as each of the other numbers. Find the probability distribution. What is the probability of rolling an even number?

Short Answer

Expert verified
The probability distribution of the weighted die is: P(1) = P(2) = \(\frac{3}{10}\), and P(3) = P(4) = P(5) = P(6) = \(\frac{1}{10}\). The probability of rolling an even number on this weighted die is \(\frac{1}{2}\), or 50%.

Step by step solution

01

Calculate the probability of each number

Let's represent the probability of rolling 1 or 2 as \(3x\), and the probability of rolling any other number as \(x\). The total probability of all six numbers should be equal to 1. We have a total of six numbers on the die: two numbers with a probability of \(3x\) and four with a probability of \(x\), so the total probability is \(2(3x) + 4(x)\), which should be equal to 1. So, we have: \(2(3x) + 4(x) = 1\) Now, we will solve for \(x\). \(6x + 4x = 1\) \(10x = 1\) \(x = \frac{1}{10}\) Now that we know the value of \(x\), we can find the probabilities of each number. Probabilities: - Rolling a 1 or 2: \(3x = 3\left(\frac{1}{10}\right) = \frac{3}{10}\) - Rolling a 3, 4, 5, or 6: \(x = \frac{1}{10}\) The probability distribution of this weighted die is: - P(1) = P(2) = \(\frac{3}{10}\) - P(3) = P(4) = P(5) = P(6) = \(\frac{1}{10}\)
02

Find the probability of rolling an even number

Now, we will find the probability of rolling an even number (2, 4, or 6). To find the probability, we add up the probabilities of rolling a 2, 4, or 6: P(even) = P(2) + P(4) + P(6) = \(\frac{3}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}\) So, the probability of rolling an even number on this weighted die is \(\frac{1}{2}\), or 50%.

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Key Concepts

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

Weighted Die
A weighted die is different from a fair die because it's designed to land on certain numbers more frequently than others. In this case, the numbers 1 and 2 are more likely to appear than the other numbers on the die.

A regular six-sided die, often called a fair die, gives equal chances for each number to show up, meaning each number has a probability of \( rac{1}{6}\). However, with a weighted die, probabilities are not equal for all outcomes.

For this weighted die, 1 and 2 appear three times as often as the other numbers. This causes an imbalance in the likelihood of rolling certain numbers compared to a fair die, resulting in a probability distribution that reflects these differences.
Probability Calculation
Calculating probabilities involves understanding the likelihood of different outcomes occurring. To calculate the probability for this weighted die, we first represent the probability of rolling a 1 or 2 as \(3x\), while rolling any other number as \(x\).

Given there are two numbers with likelihood \(3x\) and four numbers with likelihood \(x\), all probabilities add up to 1 (because one of the six possible outcomes must occur). The equation is:
  • \(2(3x) + 4(x) = 1\)
  • Solving gives \(10x = 1\), and thus \(x = \frac{1}{10}\)
With this value, individual probabilities can be determined:
  • Probability of rolling 1 or 2: \(3x = \frac{3}{10}\)
  • Probability of rolling 3, 4, 5, or 6: \(x = \frac{1}{10}\)
This calculation results in a clear representation of what to expect when rolling the die.
Even Number Probability
An even number on a die is any number that can be divided evenly by 2, such as 2, 4, and 6. Understanding the probability of rolling such numbers helps determine how likely those specific outcomes are on a weighted die.

To find this probability, we sum the probabilities of rolling each even number. Here, we use the probabilities calculated previously:
  • Probability of rolling a 2: \(\frac{3}{10}\)
  • Probability of rolling a 4: \(\frac{1}{10}\)
  • Probability of rolling a 6: \(\frac{1}{10}\)
Adding these, the probability of rolling an even number becomes:\[P(2 \text{ or } 4 \text{ or } 6) = \frac{3}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}\]This equates to a 50% chance, providing insight into how likely rolling an even number is on the weighted die.

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Most popular questions from this chapter

In exercise, you are asked to calculate the probability of being dealt various poker hands. (Recall that a poker player is dealt 5 cards at random from a standard deck of 52.) Express each of your answers as a decimal rounded to four decimal places, unless otherwise stated. Two cards with one denomination, two with another, and one with a third.

Describe the sample space \(S\) of the experiment and list the elements of the given event. (Assume that the coins are distinguishable and that what is observed are the faces or numbers that face up.) HINT [See Examples 1-3.] A sequence of two different digits is randomly chosen from the digits \(0-4\); the first digit is twice the second.

If two indistinguishable dice are rolled, what is the probability of the event \(\\{(4,4),(2,3)\\} ?\) What is the corresponding event for a pair of distinguishable dice? HINT [See Example 2.]

Find a formula for the probability of the union of three (not necessarily mutually exclusive) events \(A, B\), and \(C\).

based on the following table, which shows the results of a survey of authors by a (fictitious) publishing company. HINT [See Example 5.] $$ \begin{array}{|r|c|c|c|} \hline & \text { New Authors } & \text { Established Authors } & \text { Total } \\ \hline \text { Successful } & 5 & 25 & 30 \\ \hline \text { Unsuccessful } & 15 & 55 & 70 \\ \hline \text { Total } & 20 & 80 & 100 \\ \hline \end{array} $$ Consider the following events: \(S:\) an author is successful; \(U\) : an author is unsuccessful; \(N:\) an author is new; and \(E:\) an author is established. Describe the events \(S \cap N\) and \(S \cup N\) in words. Use the table to compute \(n(S \cap N)\) and \(n(S \cup N)\).
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