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Use the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. Six different colored dice are rolled. Of interest is the number of dice that show a one. a. In words, define the random variable \(X.\) b. List the values that \(X\) may take on. c. Give the distribution of \(X . X \sim\) _____(___,___) d. On average, how many dice would you expect to show a one? e. Find the probability that all six dice show a one. f. Is it more likely that three or that four dice will show a one? Use numbers to justify your answer numerically.

Short Answer

Expert verified
a. Number of dice showing a one. b. 0 to 6. c. Binomial(6, 1/6). d. 1 die. e. \(\frac{1}{7776}\). f. More likely for 3 dice to show a one.

Step by step solution

01

Define the Random Variable

The random variable \(X\) represents the number of dice, out of the six rolled, that land showing a one.
02

Possible Values for X

Since there are six dice, each showing a one is an event that can either happen or not. Therefore, \(X\) can take on the values \(0, 1, 2, 3, 4, 5,\) or \(6.\)
03

Specify the Distribution

The distribution of \(X\) is a binomial distribution where the number of trials \(n = 6\) (the number of dice) and the probability of success \(p = \frac{1}{6}\) (probability of a die showing a one). Thus, \(X \sim \text{Binomial}(6, \frac{1}{6}).\)
04

Calculate the Expected Value

The expected value for a binomial random variable is calculated using the formula \(E(X) = n \times p\). Here, \(n = 6\) and \(p = \frac{1}{6}\), so \(E(X) = 6 \times \frac{1}{6} = 1.\)
05

Probability All Dice Show a One

The probability that all six dice show a one, which is when \(X = 6\), is \( \left( \frac{1}{6} \right)^6 = \frac{1}{7776}.\)
06

Compare Probabilities for X=3 and X=4

Calculate the probabilities using the binomial formula: \(P(X = k) = \binom{6}{k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{6-k}\). Compute \(P(X=3)\) and \(P(X=4)\). - For \(X=3\), \(P(X=3) = \binom{6}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^3 = 20 \times \frac{1}{216} \times \frac{125}{216} \approx 0.0540.\)- For \(X=4\), \(P(X=4) = \binom{6}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^2 = 15 \times \frac{1}{1296} \times \frac{25}{36} \approx 0.0041.\)
07

Determine More Likely Outcome

Since \(P(X=3) \approx 0.0540\) and \(P(X=4) \approx 0.0041\), it is more likely to get exactly three dice showing a one than four.

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Key Concepts

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

Probability
Probability is a measure indicating how likely an event is to occur. It ranges from 0 (the event will not happen) to 1 (the event will definitely happen). In the context of the dice rolling exercise, we are interested in the probability of rolling a one with a single die, which is calculated as the number of favorable outcomes divided by the total number of possible outcomes.

For a standard six-sided die, the probability of rolling any particular number, including a one, is \( \frac{1}{6} \). This is because there is one 'one' on a die, and six possible outcomes in total. Extending this to six dice, we enter the realm of binomial probability, where we multiply this probability across multiple dice, considering combinations where a specific number of dice show a one.

Understanding basic probability helps form the foundation for approaching problems involving random variables and expected values, especially when analyzing occurrences like rolling a die.
Random Variable
A random variable is a numerical outcome of a random process. In our dice-rolling scenario, the random variable, denoted as \(X\), represents the number of dice landing on a one. Since \(X\) can vary with each roll, depending on how many dice show a one, it is termed as a random variable.

For six dice, the possible values of \(X\) range from 0 (none of the dice show a one) to 6 (all dice show a one). The values that \(X\) takes are influenced by the distribution of the process, which in this case is the binomial distribution. Each trial (or dice roll) is independent and follows the same probability, making the problem suitable for binomial analysis.

By observing the behavior of \(X\), we can gain insights into the likelihood of different scenarios, which is useful in predicting outcomes and making informed assumptions about the process at hand.
Expected Value
The expected value is essentially the average outcome you would predict after many repetitions of the same experiment. It's a fundamental concept in probability that provides the mean of a distribution based on weighted outcomes.

For the random variable \(X\), which indicates the number of dice showing a one, the expected value formula for a binomial distribution is given by \(E(X) = n \times p\), where \(n\) is the number of trials, and \(p\) is the probability of success in each trial. In this example, \(E(X) = 6 \times \frac{1}{6}\). Hence, on average, you would expect one die to show a one when six dice are rolled.

This concept helps in summarizing the central tendency of the outcomes and aids in making comparisons between different scenarios. Expected value is widely used in statistics for decision making, risk assessment, and predicting long-term results.

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Most popular questions from this chapter

Use the following information to answer the next five exercises: Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five events 35% of the time, four events 25% of the time, three events 20% of the time, two events 10% of the time, one event 5% of the time, and no events 5% of the time. Find the probability that Javier volunteers for at least one event each month. \(P(x>0)=\)________

Use the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. The chance of an IRS audit for a tax return with over \(\$ 25,000\) in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent. a. In words, define the random variable \( X\). b. List the values that \(X\) may take on. c. Give the distribution of \(X . X \sim\) ___ (___,____) d. How many audits are expected in a 20-year period? e. Find the probability that a person is not audited at all. f. Find the probability that a person is audited more than twice.

Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. On average, for every 25 patients calling in, how many do you expect to have the flu?

Use the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. According to The World Bank, only 9% of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let \(X\) = the number of people who have access to electricity. a. What is the probability distribution for \(X\)? b. Using the formulas, calculate the mean and standard deviation of \(X\). c. Use your calculator to find the probability that 15 people in the sample have access to electricity. d. Find the probability that at most ten people in the sample have access to electricity. e. Find the probability that more than 25 people in the sample have access to electricity.

Identify the mistake in the probability distribution table. $$\begin{array}{|c|c|c|}\hline x & {P(x)} & {x^{*} P(x)} \\ \hline 1 & {0.15} & {0.15} \\ \hline 2 & {0.25} & {0.40} \\ \hline 3 & {0.25} & {0.85} \\\ \hline 4 & {0.20} & {0.85} \\ \hline 5 & {0.15} & {1} \\ \hline\end{array}$$

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