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Two fair six-sided dice are tossed independently. Let \(M=\) the maximum of the two tosses [thus \(M(1,5)=5, M(3,3)=3\), etc.]. a. What is the pmf of \(M\) ? [Hint: First determine \(p(1)\), then \(p(2)\), and so on.] b. Determine the cdf of \(M\) and graph it.

Short Answer

Expert verified
The PMF is: \(p(m) = \frac{1}{36}, \frac{1}{12}, \frac{5}{36}, \frac{7}{36}, \frac{1}{4}, \frac{11}{36}\). The CDF is a step function from \(0\) to \(1\).

Step by step solution

01

Understand the Problem

To find the pmf of \(M\), the maximum of two dice rolls, we must evaluate how often each possible maximum appears. The values \(M\) can take are from 1 to 6.
02

Calculate \(p(1)\)

For \(M = 1\), both dice must show 1. The probability for this is \(p(1) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
03

Calculate \(p(2)\)

For \(M = 2\), at least one die shows a 2 while the maximum possible roll below 2 is 1. There are three favorable outcomes: \((1,2), (2,1), (2,2)\). Thus, \(p(2) = \frac{3}{36} = \frac{1}{12}\).
04

Calculate \(p(3)\)

For \(M = 3\), at least one die must show 3, and neither can show more than 3. The outcomes are \((3,1), (3,2), (3,3), (1,3), (2,3)\). So, \(p(3) = \frac{5}{36}\).
05

Calculate \(p(4)\)

For \(M = 4\), at least one die shows 4, and neither can show more than 4. Outcomes: \((4,1), (4,2), (4,3), (4,4), (1,4), (2,4), (3,4)\). Thus, \(p(4) = \frac{7}{36}\).
06

Calculate \(p(5)\)

For \(M = 5\), at least one die shows 5, and neither shows more than 5. Outcomes: \((5,1), (5,2), (5,3), (5,4), (5,5), (1,5), (2,5), (3,5), (4,5)\). Thus, \(p(5) = \frac{9}{36} = \frac{1}{4}\).
07

Calculate \(p(6)\)

For \(M = 6\), at least one die must be 6, and neither exceeds 6. Outcomes: \((6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (1,6), (2,6), (3,6), (4,6), (5,6)\). Thus, \(p(6) = \frac{11}{36}\).
08

Construct the PMF

The probability mass function \(p(m)\) is: \[\begin{align*}p(1) &= \frac{1}{36}, \p(2) &= \frac{1}{12}, \p(3) &= \frac{5}{36}, \p(4) &= \frac{7}{36}, \p(5) &= \frac{1}{4}, \p(6) &= \frac{11}{36}.\end{align*}\]
09

Determine the CDF

The cumulative distribution function (CDF) \(F(m)\) is calculated by summing probabilities:\[\begin{align*}F(1) &= \frac{1}{36}, \F(2) &= \frac{1}{36} + \frac{1}{12} = \frac{4}{36}, \F(3) &= \frac{4}{36} + \frac{5}{36} = \frac{9}{36}, \F(4) &= \frac{9}{36} + \frac{7}{36} = \frac{16}{36}, \F(5) &= \frac{16}{36} + \frac{1}{4} = \frac{25}{36}, \F(6) &= \frac{25}{36} + \frac{11}{36} = 1.\end{align*}\]The CDF values are plugged into their respective function at points \(m=1, 2, 3, 4, 5,\) and \(6\).
10

Graph the CDF

The graph of the CDF is a step function, increasing at each possible value of \(M\). The horizontal axis represents \(M\) values from 1 to 6, while the vertical axis represents their cumulative probabilities from \(\frac{1}{36}\) to 1.

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Key Concepts

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

Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is an essential concept in probability that tells us the probability that a random variable is less than or equal to a certain value. In simpler terms, it helps us understand the accumulation of probabilities up to a specific point within a probability distribution.
For the random variable in our exercise, which is the maximum roll of two dice, the CDF is a way to show, for each possible maximum outcome, the summed probabilities of observing this maximum or any lower value.
The CDF of the maximum roll on two dice, denoted as \( F(m) \), is calculated by adding up the probabilities from the Probability Mass Function (PMF) up to the value \( m \).
For example:
  • \( F(1) = \frac{1}{36} \)
  • \( F(2) = \frac{1}{36} + \frac{1}{12} = \frac{4}{36} \)
  • And so on, until \( F(6) = 1 \), indicating that you will always roll a number between 1 and 6
The CDF is particularly useful for understanding probabilities in a cumulative manner and is often depicted as a step function, showcasing the sum of the PMF values as you move up the possible outcomes of a discrete random variable.
Probability Mass Function
The Probability Mass Function (PMF) is a function that provides the probabilities of all possible discrete outcomes of a random variable. It essentially maps each outcome to its likelihood, which is essential in understanding the probability distribution of discrete random variables like rolling dice.
In our exercise, the PMF describes the probability of the maximum number shown when two six-sided dice are rolled. This is given by \( p(m) \), where \( m \) is the possible maximum value:
  • \( p(1) = \frac{1}{36} \)
  • \( p(2) = \frac{1}{12} \)
  • \( p(3) = \frac{5}{36} \)
  • \( p(4) = \frac{7}{36} \)
  • \( p(5) = \frac{1}{4} \)
  • \( p(6) = \frac{11}{36} \)
These probabilities add up to 1, as they encompass all possible maximum outcomes when two dice are rolled. The PMF is vital for clearly specifying the likelihood of each possible outcome, allowing us to further compute cumulative probabilities and draw probability-related insights.
Discrete Random Variable
A Discrete Random Variable is a type of variable that can take on a finite or countable number of values, each associated with a probability. They are the cornerstone of probability theory when dealing with real-world applications like games involving dice or cards.
In our exercise, the discrete random variable is \( M \), representing the maximum outcome when two dice are rolled.
Since each die can show a number between 1 to 6, the possible values for \( M \) range from 1 to 6 as well.
  • \( M = 1 \) indicates both dice show a 1
  • \( M = 2 \) occurs if at least one die shows a 2, and neither die shows more than 2
  • ...continuing up to \( M = 6 \), which is the highest possible maximum outcome
This highlights how discrete random variables quantify outcomes in scenarios where results are distinct and separate, forming the basis for more complex probability measures like the PMF and CDF.

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Most popular questions from this chapter

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