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Suppose a die is rolled twice. What are the possible values that the following random variables can take on? (a) The maximum value to appear in the two rolls. (b) The minimum value to appear in the two rolls. (c) The sum of the two rolls. (d) The value of the first roll minus the value of the second roll.

Short Answer

Expert verified
Possible values for each random variable: (a) Maximum value of the two rolls: {1, 2, 3, 4, 5, 6} (b) Minimum value of the two rolls: {1, 2, 3, 4, 5, 6} (c) Sum of the two rolls: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} (d) Difference of the two rolls (first roll - second roll): {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

Step by step solution

01

(a) Finding the maximum value of the two rolls

To find the possible maximum values, first list all possible combinations of the two rolls. There are 6 faces and we roll the die twice, giving us a total of 6x6 = 36 combinations. Now, for each combination, find the maximum value and list all unique values. Note: Since we are concerned only with the unique possible values of the maximum value, it's not necessary to list all 36 combinations. We can simply compare each number from 1 to 6 with all other numbers, finding their maximum values and listing them. 1 1 1 2 1 3 1 4 1 5 1 6 2 2 2 3 ... 6 6 Possible maximum values: {1, 2, 3, 4, 5, 6}
02

(b) Finding the minimum value of the two rolls

Now, we will find the possible minimum values of the two rolls. Using the same combinations and approach as before, this time find the minimum value for each combination and list all unique values. Possible minimum values: {1, 2, 3, 4, 5, 6}
03

(c) Finding the sum of the two rolls

Next, we will find the possible values of the sum of the two rolls. For this, we actually need to consider all 36 combinations, find the sum for each and list all unique values. List all combinations and their sums: 1+1=2 1+2=3 1+3=4 ... 6+6=12 Possible sum values: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
04

(d) Finding the difference of the two rolls (first roll - second roll)

Finally, we will find the possible values for the difference of the two rolls (first roll minus the second roll). Use the same approach as before, listing all combinations and finding the difference for each, while listing all unique values. 1-1=0 1-2=-1 1-3=-2 ... 6-6=0 Possible difference values: {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} For each of the given random variables, we have determined the possible values based on all possible combinations of rolling two dice.

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Most popular questions from this chapter

A coin having probability \(p\) of coming up heads is successively flipped until the \(r\) th head appears. Argue that \(X\), the number of flips required, will be \(n, n \geq r\), with probability $$ P[X=n\\}=\left(\begin{array}{c} n-1 \\ r-1 \end{array}\right) p^{T}(1-p)^{n-r}, \quad n \geq r $$ This is known as the negative binomial distribution. Hint: How many successes must there be in the first \(n-1\) trials?

Let \(X\) be binomially distributed with parameters \(n\) and \(p\). Show that as \(k\) goes from 0 to \(n, P(X=k)\) increases monotonically, then decreases monotonically reaching its largest value (a) in the case that \((n+1) p\) is an integer, when \(k\) equals either \((n+1) p-1\) or \((n+1) p\) (b) in the case that \((n+1) p\) is not an integer, when \(k\) satisfies \((n+1) p-1

If the distribution function of \(F\) is given by $$ F(b)=\left\\{\begin{array}{ll} 0, & b<0 \\ \frac{1}{2}, & 0 \leq b<1 \\ \frac{3}{3}, & 1 \leq b<2 \\ \frac{4}{3}, & 2 \leq b<3 \\ \frac{9}{10}, & 3 \leq b<3.5 \\ 1, & b \geq 3.5 \end{array}\right. $$ calculate the probability mass function of \(X\).

Suppose that the joint probability mass function of \(X\) and \(Y\) is $$ P(X=i, Y=j)=\left(\begin{array}{l} j \\ i \end{array}\right) e^{-2 \lambda} \lambda^{i} / j !, \quad 0 \leq i \leq j $$ (a) Find the probability mass function of \(Y\). (b) Find the probability mass function of \(X\). (c) Find the probability mass function of \(Y-X\).

Let \(X\) and \(Y\) each take on either the value 1 or \(-1\). Let $$ \begin{aligned} p(1,1) &=P\\{X=1, Y=1\\} \\ p(1,-1) &=P[X=1, Y=-1\\} \\ p(-1,1) &=P[X=-1, Y=1\\} \\ p(-1,-1) &=P\\{X=-1, Y=-1\\} \end{aligned} $$ Suppose that \(E[X]=E[Y]=0\). Show that (a) \(p(1,1)=p(-1,-1) ;\) (b) \(p(1,-1)=p(-1,1)\). Let \(p=2 p(1,1) .\) Find (c) \(\operatorname{Var}(X)\); (d) \(\operatorname{Var}(Y)\) (e) \(\operatorname{Cov}(X, Y)\).

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