91Ó°ÊÓ

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to land face up as any particular odd-numbered face. Consider the chance experiment that consists of rolling this die. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p, P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p\) Now use a condition on the sum of these probabilities to determine \(p\).) b. What is the probability that the number showing is an odd number? at most three? c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots\), \(P\left(O_{6}\right)=6 c\). What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

Step by step solution

01

Identifying the probabilities of the six simple events

Let's denote the probability of landing an odd face as \(p\), therefore the probability of landing an even face is \(2p\). According to the rule of Total Probability, the sum of probabilities for all possible outcomes should equal 1. Therefore, for our loaded die, where we have 3 odd faces and 3 even faces, we derive the equation: \(3p + 3 \cdot 2p = 1\), which simplifies to \(9p = 1\). Solving for \(p\) yields \(p = 1/9\). Therefore, the probability for each face would be \(P(O_1) = P(O_3) = P(O_5) = 1/9\) and \(P(O_2) = P(O_4) = P(O_6) = 2/9\).

02

Probabilities of rolling an odd number or a number at most three

The probability of rolling an odd number would be the sum of probabilities for faces 1,3 and 5, totaling \( P(O_1) + P(O_3) + P(O_5) = 1/3\). Similarly, the probability of rolling a number at most three would be the sum of probabilities for faces 1, 2, and 3, adding up to \(P(O_1) + P(O_2) + P(O_3) = 4/9.\)

03

New conditions for the simple events

The third part of the problem supposes that the probability of each event is proportional to the number showing on the upturned face. Let's denote the proportional constant as c. The total sum of probabilities still must be 1, which gives us the equation: \(c\cdot1 + c\cdot2 + c\cdot3 + c\cdot4 + c\cdot5 + c\cdot6 = 1 \), simplifying to \(c \cdot 21 = 1\). So, we obtain \(c = 1/21\). Now, the probabilities for each face would be \(P(O_1) = c, P(O_2) = 2c, ..., P(O_6) = 6c\)

04

New probabilities of rolling an odd number or a number at most three

With the new probabilities, the chance of rolling an odd number becomes \(P(O_1) + P(O_3) + P(O_5) = 9/21 = 3/7\), and the chance of rolling a number at most three is \(P(O_1) + P(O_2) + P(O_3) = 6/21 = 2/7).\n

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Key Concepts

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

Loaded Die

A loaded die is one that is intentionally weighted to favor certain outcomes over others. In the case of our exercise, the die is "loaded" so that even-numbered outcomes are twice as likely as odd-numbered outcomes.
This means that while a standard fair die would have each face showing up with equal probability, a loaded die skews these probabilities.
If you imagine a balanced die, each face appears 1/6 of the time. But here, we assign probabilities based on loading instructions, which affects how likely each number is to show up when rolled. This manipulation of likelihoods is what makes a die "loaded."
In probability problems like this, understanding how the die is loaded helps determine the probability of each simple event occurring.

Simple Events

A simple event in probability refers to an individual outcome of an experiment. For our exercise, each face of the die represents a simple event when rolled.
When classifying the events, we use the notation:

• \(O_1\) for rolling a 1,
• \(O_2\) for rolling a 2,
• \(O_3\) for rolling a 3,
• and so on until \(O_6\), rolling a 6.

Each of these represents a distinct simple event, and understanding this helps in calculating their respective probabilities.
In loaded die scenarios, these probabilities aren't equal, so each event must be analyzed separately based on the conditions given.

Rule of Total Probability

The rule of total probability states that the total of all possible outcomes' probabilities must equal 1. This is essential for constructing probability distributions in probabilistic models.
In our problem, even though the die is loaded, the basic rule doesn't change:

• Each face of the six-sided die has a probability, and when you sum up these individual probabilities, they should equal 1.

For a loaded die, we've learned that:

• For odd faces, the probability is \(p\),
• and for even faces, it's \(2p\).

Thus, knowing there are 3 odd and 3 even faces, the equation becomes \(3p + 3(2p) = 1\), leading us to solve that \(p = 1/9\).
This calculation confirms the rule of total probability, ensuring the sum remains constant.

Proportional Probability

Proportional probability is a scenario where probabilities are assigned based on a ratio or weight relative to a reference value or condition.
In the latter part of our exercise, the probability of each event is directly proportional to the face value of the die. This means:

• Face 1 has a probability of \(c\),
• Face 2 has \(2c\),
• Face 3 has \(3c\),
and so forth until

Face 6 with \(6c\). These probabilities are calculated by ensuring they sum up to 1, adhering to the rule of total probability.
The equation becomes \(c(1 + 2 + 3 + 4 + 5 + 6) = 1\), which simplifies to \(c = 1/21\).
By assigning these proportional probabilities, we craft a probability distribution that reflects the likelihood of each outcome based on the load configuration and face value.