In: Math

Suppose you are rolling two independent fair dice. You may have one of the following outcomes

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,2) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now define a random variable Y = the absolute value of the
difference of the two numbers

a. Complete the following pmf of Y with necessary calculations and
reasoning.

b. Find the mgf of Y

c. Now further consider another random variable X = the sum of the two numbers. Do you think X and Y are independent? Briefly explain your reasons

Following is the sample space when we roll two fair dice:

Above sample space shows the absolute difference of all the outcomes after = sign. So pmf of Y will be

Y | P(Y) | P(Y) |

0 | 6/36 | 0.17 |

1 | 10/36 | 0.28 |

2 | 8/36 | 0.22 |

3 | 6/36 | 0.17 |

4 | 4/36 | 0.11 |

5 | 2/36 | 0.06 |

Total | 1 |

And following table shows the calculations for expected values:

Y | P(Y) | P(Y) | YP(Y) |

0 | 6/36 | 0.17 | 0 |

1 | 10/36 | 0.28 | 0.28 |

2 | 8/36 | 0.22 | 0.44 |

3 | 6/36 | 0.17 | 0.51 |

4 | 4/36 | 0.11 | 0.44 |

5 | 2/36 | 0.06 | 0.3 |

Total | 1 | 1.97 |

(b)

The MGF of Y is

Hence,

(c)

Following is the possible outcomes when we sum the outcomes and absolute difference of rolls of two dice:

In each outcome (a,b)=c,d : a shows outcome of first die, b shows outcome of second die and c shows sum c =a+b and d shows abs(a-b).

Following table shows the joint pdf and marginal pdfs of X and Y

Y | ||||||||

0 | 1 | 2 | 3 | 4 | 5 | P(X=x) | ||

2 | 1/36 | 0 | 0 | 0 | 0 | 0 | 1/36 | |

3 | 0 | 2/36 | 0 | 0 | 0 | 0 | 2/36 | |

4 | 1/36 | 0 | 2/36 | 0 | 0 | 0 | 3/36 | |

5 | 0 | 2/36 | 0 | 2/36 | 0 | 0 | 4/36 | |

X | 6 | 1/36 | 0 | 2/36 | 0 | 2/36 | 0 | 5/36 |

7 | 0 | 2/36 | 0 | 2/36 | 0 | 2/36 | 6/36 | |

8 | 1/36 | 0 | 2/36 | 0 | 2/36 | 0 | 5/36 | |

9 | 0 | 2/36 | 0 | 2/36 | 0 | 0 | 4/36 | |

10 | 1/36 | 0 | 2/36 | 0 | 0 | 0 | 3/36 | |

11 | 0 | 2/36 | 0 | 0 | 0 | 0 | 2/36 | |

12 | 1/36 | 0 | 0 | 0 | 0 | 0 | 1/36 | |

Y | 6/36 | 10/36 | 8/36 | 6/36 | 4/36 | 2/36 | 1 |

Following table shows the calculations for E(XY):

X | Y | P(X=x, Y=y) | xy*P(X=x, Y=y) |

2 | 0 | 1/36 | 0/36 |

2 | 1 | 0 | 2/36 |

2 | 2 | 0 | 4/36 |

2 | 3 | 0 | 6/36 |

2 | 4 | 0 | 8/36 |

2 | 5 | 0 | 10/36 |

3 | 0 | 0 | 0/36 |

3 | 1 | 2/36 | 3/36 |

3 | 2 | 0 | 6/36 |

3 | 3 | 0 | 9/36 |

3 | 4 | 0 | 12/36 |

3 | 5 | 0 | 15/36 |

4 | 0 | 1/36 | 0/36 |

4 | 1 | 0 | 4/36 |

4 | 2 | 2/36 | 8/36 |

4 | 3 | 0 | 12/36 |

4 | 4 | 0 | 16/36 |

4 | 5 | 0 | 20/36 |

5 | 0 | 0 | 0/36 |

5 | 1 | 2/36 | 5/36 |

5 | 2 | 0 | 10/36 |

5 | 3 | 2/36 | 15/36 |

5 | 4 | 0 | 20/36 |

5 | 5 | 0 | 25/36 |

6 | 0 | 1/36 | 0/36 |

6 | 1 | 0 | 6/36 |

6 | 2 | 2/36 | 12/36 |

6 | 3 | 0 | 18/36 |

6 | 4 | 2/36 | 24/36 |

6 | 5 | 0 | 30/36 |

7 | 0 | 0 | 0/36 |

7 | 1 | 2/36 | 7/36 |

7 | 2 | 0 | 14/36 |

7 | 3 | 2/36 | 21/36 |

7 | 4 | 0 | 28/36 |

7 | 5 | 2/36 | 35/36 |

8 | 0 | 1/36 | 0/36 |

8 | 1 | 0 | 8/36 |

8 | 2 | 2/36 | 16/36 |

8 | 3 | 0 | 24/36 |

8 | 4 | 2/36 | 32/36 |

8 | 5 | 0 | 40/36 |

9 | 0 | 0 | 0/36 |

9 | 1 | 2/36 | 9/36 |

9 | 2 | 0 | 18/36 |

9 | 3 | 2/36 | 27/36 |

9 | 4 | 0 | 36/36 |

9 | 5 | 0 | 45/36 |

10 | 0 | 1/36 | 0/36 |

10 | 1 | 0 | 10/36 |

10 | 2 | 2/36 | 20/36 |

10 | 3 | 0 | 30/36 |

10 | 4 | 0 | 40/36 |

10 | 5 | 0 | 50/36 |

11 | 0 | 0 | 0/36 |

11 | 1 | 2/36 | 11/36 |

11 | 2 | 0 | 22/36 |

11 | 3 | 0 | 33/36 |

11 | 4 | 0 | 44/36 |

11 | 5 | 0 | 55/36 |

12 | 0 | 1/36 | 0/36 |

12 | 1 | 0 | 12/36 |

12 | 2 | 0 | 24/36 |

12 | 3 | 0 | 36/36 |

12 | 4 | 0 | 48/36 |

12 | 5 | 0 | 60/36 |

Total | 1155/36 |

SO,

E(XY) = 490 /36

---------

Following is the possible outcomes when we sum the outcomes of rolls of two dice:

Let X shows the sum of two fair die. So X can take values 2, 3, 4,...12. Following table shows the probability distribution of X:

X | P(X=x) |

2 | 1/36 |

3 | 2/36 |

4 | 3/36 |

5 | 4/36 |

6 | 5/36 |

7 | 6/36 |

8 | 5/36 |

9 | 4/36 |

10 | 3/36 |

11 | 2/36 |

12 | 1/36 |

Following table shows the calculations for mean and variance of X:

X | P(X=x) | P(X=x) | XP(X=x) | X^2*P(X) |

2 | 1/36 | 0.027778 | 0.05555556 | 0.11111111 |

3 | 2/36 | 0.055556 | 0.16666667 | 0.5 |

4 | 3/36 | 0.083333 | 0.33333333 | 1.33333333 |

5 | 4/36 | 0.111111 | 0.55555556 | 2.77777778 |

6 | 5/36 | 0.138889 | 0.83333333 | 5 |

7 | 6/36 | 0.166667 | 1.16666667 | 8.16666667 |

8 | 5/36 | 0.138889 | 1.11111111 | 8.88888889 |

9 | 4/36 | 0.111111 | 1 | 9 |

10 | 3/36 | 0.083333 | 0.83333333 | 8.33333333 |

11 | 2/36 | 0.055556 | 0.61111111 | 6.72222222 |

12 | 1/36 | 0.027778 | 0.33333333 | 4 |

Total | 1 | 1 | 7 | 54.8333333 |

The expected value of X is

The variance of X:

Var(X) = 54.8333 - 7^{2} = 5.8333

-------------------------------------------

Following table shows the calculations for expected values:

Y | P(Y) | P(Y) | YP(Y) |

0 | 6/36 | 0.17 | 0 |

1 | 10/36 | 0.28 | 0.28 |

2 | 8/36 | 0.22 | 0.44 |

3 | 6/36 | 0.17 | 0.51 |

4 | 4/36 | 0.11 | 0.44 |

5 | 2/36 | 0.06 | 0.3 |

Total | 1 | 1.97 |

So expected value of X is

Following table shows the calculations for variance of Y:

Y | P(Y) | YP(Y) | y^2P(y) |

0 | 0.17 | 0 | 0 |

1 | 0.28 | 0.28 | 0.28 |

2 | 0.22 | 0.44 | 0.88 |

3 | 0.17 | 0.51 | 1.53 |

4 | 0.11 | 0.44 | 1.76 |

5 | 0.06 | 0.3 | 1.5 |

Total | 1 | 1.97 | 5.95 |

The expected value is

E(Y) = sum xP(X=x) = 1.97

So,

----------------------------

The co-variance is

Cov(X,Y) = E(XY) - E(X)E(Y) = (490/36) - (7*1.97) = -0.1789

The correlation coefficient is

Corr(X,Y) = Cov(X,Y) / (sqrt(Var(X) * Var(Y)) = -0.1789 / (sqrt( 5.8333 * 2.0691)) = - 0.0515

Since correlation coefficient is not zero so X ad Y are not independent.

Consider rolling a fair dice. You keep rolling the dice until
you see all of the faces (from number 1 to 6) at least once. What
is your expected number of rolls?

Rolling doubles When rolling two fair, 6-sided
dice, the probability of rolling doubles is 1/6. Suppose Elias
rolls the dice 4 times. Let W = the number of times he
rolls doubles. The probability distribution of W is shown
here. Find the probability that Elias rolls doubles more than
twice.
Value
0
1
2
3
4
Probability
0.482
0.386
0.116
0.015
0.001

Let X be the outcome of rolling a fair six-sided dice. The
possible outcomes or X are 1,2,3,4,5 and 6 and all are equally
likely. What is the cumulative distribution function F(x)?

An experiment is rolling two fair dice and adding the spots
together. Find the following probabilities; enter all
answers as simplified fractions using the / bar between numerator
and denominator, with no extra space
Blank #1: Find the probability of getting a sum
of 3.
Blank #2: Find the probability of getting the
first die as a 4.
Blank #3: Find the probability of getting a sum
of 8.
Blank #4: Find the probability of getting a sum
of 3...

Q\ Suppose we have dice; the dice have six outcomes find the
probability of the
following: -
a) What’s the probability of rolling a 4?
b) What’s the probability of rolling an odd number?
c) What’s the probability of rolling less than 6?
d) What’s the probability of rolling a 1 and 3?
e) What’s the probability of rolling a 0?
f) The probability of event a or c?
g) What’s the probability of rolling not event b?

11.)
An experiment consists of rolling two fair dice and adding the
dots on the two sides facing up. Assuming each simple event is as
likely as any other, find the probability that the sum of the dots
is greater than 2.
The probability that the sum of the dots is greater than 2
is
12.)
An experiment consists of rolling two fair dice and adding the
dots on the two sides facing up. Find the probability of the sum...

Q1. Two fair dice are rolled. What is the probability
of…
a) Rolling a sum
of 4 or doubles?
b) Rolling a sum of 4 and doubles
c) Rolling a sum
of 2, 4 times in a row?
Q2. True or False. A discrete sample space is one in
which outcomes are counted

suppose he roll two dice one black and one red you record the
outcomes in order for example the outcome (35) notes at three on
the black die And a five on the red dye. What a beauty event that
the black guy is even, B be the event at the red dye is odd, and C
be the event that the dice sum to 10
(a) List and describe the outcomes in the sample space S and
the events...

Two fair dice are rolled. What is the probability of…
a)Rolling a total of 8?
b) Rolling a total greater than 5?
c)Rolling doubles?
d)Rolling a sum of 6 or a sum of 8?
e)Rolling a sum of 4 or doubles?
f)Rolling a sum of 4 and doubles?
g)Rolling a sum of 2, 4 times in a row?

When you have two dice rolling at the same time, find the
following probabilities with proper explanations.
(1) When you will observe a total of 8
(2) When you will observe a total of 4 or a total of 6
(3) When you will observe at least a total of 2

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