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Loaded Dice. There are many ways to produce crooked dice. To load a die so that 6 comes up too often and 1 (which is opposite 6) comes up too seldom, add a bit of lead to the filling of the spot on the 1 face. If a die is loaded so that 6 comes up with probability \(0.2\) and the probabilities of the \(2,3,4\), and 5 faces are not affected, what is the assignment of probabilities to the six faces?

Short Answer

Expert verified
Face '1' = 0.133, Face '6' = 0.2, Faces '2', '3', '4', '5' = 0.1667 each.

Step by step solution

01

Identify Given Data

We are given that a loaded die has a probability of landing on the face with '6' as 0.2. We also know that the probabilities of the faces 2, 3, 4, and 5 remain unchanged. A fair die would give each face a probability of \( \frac{1}{6} \approx 0.1667 \).
02

Determine Probability of Face '1'

Since the die is loaded such that 6 appears more often and 1 less often, reduce the probability of the '1' face to accommodate the increased probability of the '6' face, keeping the sum of probabilities equal to 1.
03

Calculate Remaining Probabilities

Let the probability of face '1' be \( p_1 \). Since '6' has a probability of 0.2, the sum of probabilities of all faces is \( p_1 + 0.2 + 4 \times \frac{1}{6} = 1 \).
04

Solve for Unknown Probability

Substitute the known value to solve for \( p_1 \):\[ p_1 + 0.2 + \frac{4}{6} = 1 \]Simplifying the fraction gives \( \frac{4}{6} = 0.6667 \approx 0.667\). Therefore:\[ p_1 + 0.2 + 0.667 = 1 \]\[ p_1 = 1 - 0.2 - 0.667 \]\[ p_1 = 0.133 \]
05

Verify Total Probability

Ensure all probabilities sum up to 1. The assignments should be: Face '1' = 0.133, Face '6' = 0.2, and Faces '2', '3', '4', and '5' each = 0.1667. Sum all these probabilities to verify:\[ 0.133 + 0.2 + 4(0.1667) = 1 \]Calculating gives 1, confirming the probabilities are correct.

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Key Concepts

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

Loaded Dice
A loaded die, also known as a crooked die, is a type of die that has been altered in some way to change the outcome probabilities of the faces. This is often done to either increase the probability of a certain number or decrease the probability of another. In our example, extra weight is added to the side opposite face '6', specifically face '1'. This increases the likelihood of rolling a '6', while face '1' becomes less likely to appear, effectively making the die "loaded". To understand the mechanics of loading a die, consider that in a fair die, each face has an equal probability of landing face-up. By introducing some physical changes, such as adding a small amount of lead to the die on a specific face, the balance is altered, changing how the die behaves when rolled.
Fair Die
A fair die is one where each of the six faces has an equal probability of landing face-up. For a standard six-sided die, this probability is \( \frac{1}{6} \) or approximately 0.1667 for each face. This means that, on average, each face is equally likely to appear over a large number of rolls.The fairness of a die is crucial in games of chance, ensuring that each outcome is equally likely and randomly determined. This randomness and equality are what make a die "fair." If any face has a different probability than \( \frac{1}{6} \) without intentional or accidental intervention, the die would no longer be considered fair.
Assigning Probabilities
Assigning probabilities to the faces of a die involves determining how likely each face will land face-up in a roll. For a loaded die, the process starts with establishing which face is being favored—in this case, face '6' with a probability of 0.2.To maintain the sum of probabilities at 1, adjustments must be made to the probabilities of other faces. Since the die is altered, face '1' will have a reduced probability, calculated as follows:
  • The sum of probabilities must remain 1.
  • Subtract the increased probability of face '6' from 1.
  • Take the adjusted probabilities of faces '2', '3', '4', and '5', which still remain close to the fair die probability of \(0.1667\), and solve for the probability of face '1'.
  • This calculation involves ensuring that the sum of the probabilities accounts for the increased probability of '6' and the unchanged probabilities of the other faces.
Ultimately, this method ensures that probabilities are equitably reassigned while considering the impact of a loaded side on the die's overall probability distribution.
Sum of Probabilities
The sum of probabilities for all possible outcomes must always equal 1. This is because it encapsulates the certainty that one of the possible outcomes will occur. In the context of dice, this means that the cumulative probability of rolling any one face from 1 to 6 should be exactly 1. When dealing with a loaded die, special care must be taken to ensure that any adjustments to individual probabilities still retain this total sum. Specifically, though one face might have a higher or lower probability than usual, the total must still be the sum of 1. For example:
  • Face '6' is 0.2 (loaded),
  • Face '1' is recalculated to be 0.133,
  • The remaining faces '2', '3', '4', and '5' each have a probability of approximately 0.1667, as they are not affected by the loading.
By confirming the sum (0.2 + 0.133 + 4(0.1667)=1), we establish that the rule of total probability is met, validating our distribution. This checking reinforces the integrity and correctness of probability assignments.

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Most popular questions from this chapter

Running a Mile. A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean \(7.11\) minutes and standard deviation \(0.74\) minute. 12 Choose a student at random from this group and call his time for the mile \(Y\). a. Is \(Y\) a finite or continuous random variable? Explain your answer. b. Say in words what the meaning of \(P(Y \geq 8)\) is. What is this probability? c. Write the event "the student could run a mile in less than six minutes" in terms of values of the random variable \(Y\). What is the probability of this event?

Winning the ACC Tournament. The annual Atlantic Coast Conference men's basketball tournament has temporarily taken Joe's mind off of the Cleveland Indians. He says to himself, "I think that Louisville has probability \(0.05\) of winning. North Carolina's probability is twice Louisville's, and Duke's probability is four times Louisville's." a. What are Joe's personal probabilities for North Carolina and Duke? b. What is Joe's personal probability that one of the 15 teams other than Louisville, North Carolina, and Duke will win the tournament?

A Taste Test. A tea-drinking Canadian friend of yours claims to have a very refined palate. She tells you that she can tell if, in preparing a cup of tea, milk is first added to the cup and then hot tea poured into the cup or the hot tea is first poured into the cup and then the milk is added. \(1 .\) To test her claims, you prepare six cups of tea. Three have the milk added first and the other three the tea first. In a blind taste test, your friend tastes all six cups and is asked to identify the three that had the milk added first. a. How many different ways are there to select three of the six cups? (Hint: See Example 12.8, page 281.) b. If your friend is just guessing, what is the probability that she correctly identifies the three cups with the milk added first?

Probability Models? In each of the following situations, state whether or not the given assignment of probabilities to individual outcomes is legitimate- that is, satisfies the rules of probability. Remember, a legitimate model need not be a practically reasonable model. If the assignment of probabilities is not legitimate, give specific reasons for your answer. a. Roll a six-sided die and record the count of spots on the upface: $$ \begin{array}{lll} P(1)=0 & P(2)=1 / 6 & P(3)=1 / 3 \\ P(4)=1 / 3 & P(5)=1 / 6 & P(6)=0 \end{array} $$ b. Deal a card from a shuffled deck: $$ \begin{array}{rlrl} P(\text { clubs }) & =12 / 52 & P(\text { diamonds }) & =12 / 52 \\ P(\text { hearts }) & =12 / 52 & P(\text { spades }) & =16 / 52 \end{array} $$ c. Choose a college student at random and record sex and enrollment status: $$ \begin{array}{rlrl} P(\text { female full-time }) & =0.56 & P(\text { male full-time }) & =0.44 \\\ P(\text { female part-time }) & =0.24 & P(\text { male part-time }) & =0.17 \end{array} $$

Survey Accuracy. A sample survey contacted an SRS of 2220 registered voters shortly before the 2016 presidential election and asked respondents whom they planned to vote for. Election results show that \(46 \%\) of registered voters voted for Donald Trump. The proportion of the sample who voted for Trump varies, depending on which 2220 voters are in the sample. We will see later that in this situation, if we consider all possible samples of 2220 voters, the proportion of voters in each sample who planned to vote for Trump (call it \(V\) ) has approximately the Normal distribution with mean \(\mu=0.46\) and standard deviation \(\sigma=0.011\). a. If the respondents answer truthfully, what is \(P(0.44 \leq V \leq 0.48)\) ? This is the probability that the sample proportion \(V\) estimates the population proportion \(0.46\) within plus or minus \(0.02\). b. In fact, \(43 \%\) of the respondents in the actual sample said they planned to vote for Donald Trump. If respondents answer truthfully, what is \(P(V \geq 0.43)\) ?

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