Question

1a) Let an experiment consist of rolling three standard 6-sided dice.

i) Compute the expected value of the sum of the rolls.
ii) Compute the variance of the sum of the rolls.
iii) If X represents the maximum value that appears in the two rolls, what is the expected value of X?

1b) Consider an experiment where a fair die is rolled repeatedly until the first time a 3 is observed.
  
i) What is the sample space for this experiment? What is the probability that the die turns up a 3 after i rolls?
ii) What is the expected number of times we roll the die?
iii) Let E be the event that the first time a 3 turns up is after an even number of rolls. What set of outcomes belong to this event? What is the probability that Eoccurs?

Answer 1

1)

i)

sum will range from(1+1+1) 3 to (6+6+6) = 18

probabilities of sum is given below

P(3) = 1/216

P(4) = 3/216

P(5) = 6/216

P(6) = 10/216

P(7) = 15/216

P(8) = 21/216

P(9) = 25/216

P(10) = 27/216

P(11) = 27/216

P(12) = 25/216

P(13) = 21/216

P(14) = 15/216

P(15) = 10/216

P(16) = 6/216

P(17) = 3/216

P(18) = 1/216

X	P(X)	X*P(X)	X² * P(X)
3	0.0046	0.013889	0.041667
4	0.0139	0.055556	0.222222
5	0.0278	0.138889	0.694444
6	0.0463	0.277778	1.666667
7	0.0694	0.486111	3.402778
8	0.0972	0.777778	6.222222
9	0.1157	1.041667	9.375
10	0.1250	1.25	12.5
11	0.1250	1.375	15.125
12	0.1157	1.388889	16.66667
13	0.0972	1.263889	16.43056
14	0.0694	0.972222	13.61111
15	0.0463	0.694444	10.41667
16	0.0278	0.444444	7.111111
17	0.0139	0.236111	4.013889
18	0.0046	0.083333	1.5

	P(X)	X*P(X)	X² * P(X)
total sum =	1	10.5	119

mean = E[X] = Σx*P(X) =            10.5

ii)

E [ X² ] = ΣX² * P(X) =            119
          
variance = E[ X² ] - (E[ X ])² =            8.75
iii)

Following is the sample space of the rolling of two fair dice:

There are total 36 outcomes in sample space. PMF of X is :

X	Number of outcomes	P(X)
1	1	1/36
2	3	3/36
3	5	5/36
4	7	7/36
5	9	9/36
6	11	11/36
Total		1

X	P(X)	X*P(X)
1	0.0278	0.027778
2	0.0833	0.166667
3	0.1389	0.416667
4	0.1944	0.777778
5	0.2500	1.25
6	0.3056	1.833333

	P(X)	X*P(X)
total sum =	1	4.472222

mean = E[X] = Σx*P(X) =            4.4722

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for three rolls of dice

P(X=1) = 1/216

P(X=2)=7/216

P(X=3)=19/216

P(X=4)=37/216

P(X=5)=61/216

P(X=6)=91/216

hence, the expected value is:

1*1/216+2*7/216+3*19/216+4*37/216+5*61/216+6*91/216=119/24

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