Question

Two rolls of a fair die. Let x and y be the results of the two rolls, and let z = x + y.

(a) Find P[x = 4, y = 3], P[x > 3, y = 5].

(b) Find P[z = 7], P[z = 5], P[z = 3].

(c) Find the probability that at least one 6 appears given that z = 8.

(d) Find the probability that x = 6 given that z > 8.

(e) Find the probability that z = 7 given that at least one 4 was rolled. (f) Find the probability that z > 7 given that y = 4.

Answer 1

Following is the event space for the result of the two rolls of a fair die(x: first roll; y: second roll; z=x+y

#	x:First Roll	y:Second roll	z=x+y
1	1	1	2
2	1	2	3
3	1	3	4
4	1	4	5
5	1	5	6
6	1	6	7
7	2	1	3
8	2	2	4
9	2	3	5
10	2	4	6
11	2	5	7
12	2	6	8
13	3	1	4
14	3	2	5
15	3	3	6
16	3	4	7
17	3	5	8
18	3	6	9
19	4	1	5
20	4	2	6
21	4	3	7
22	4	4	8
23	4	5	9
24	4	6	10
25	5	1	6
26	5	2	7
27	5	3	8
28	5	4	9
29	5	5	10
30	5	6	11
31	6	1	7
32	6	2	8
33	6	3	9
34	6	4	10
35	6	5	11
36	6	6	12

Total number of events = 36

(a)

P[x=4, y=3] ;

From the above table;

Rows that favor x=4 , y=3 is

#	x:First Roll	y:Second roll	z=x+y
21	4	3	7

Number of event favoring Event (x=4, y=3 )= 1

P(x=4, y=3) = Number of event favoring Event (x=4, y=3 ) / Total number of events = 1/36

P[x=4, y=3] = 1/36

P[x > 3, y = 5]

From the table, rows that favor : x>3, y=5 are

#	x:First Roll	y:Second roll	z=x+y
23	4	5	9
29	5	5	10
35	6	5	11

Number of events favoring [x>3, y=5] =3

P[x>3,y=5] = 3/36 = 1/12

P[x>3,y=5] = 1/12

(b)

P[z = 7]

From the table ,

Rows favoring z=7 are :

#	x:First Roll	y:Second roll	z=x+y
6	1	6	7
11	2	5	7
16	3	4	7
21	4	3	7
26	5	2	7
31	6	1	7

Number of events favoring [z=7 ] = 6

P[z=7] = 6/36 = 1/6

P[z = 5]

From the table events favoring z=5 are

#	x:First Roll	y:Second roll	z=x+y
4	1	4	5
9	2	3	5
14	3	2	5
19	4	1	5

Number of events favoring [z=5 ] = 4

P[z=5] = 4/36 = 1/9

P[z = 3]

From the table , events favoring z=3 are

#	x:First Roll	y:Second roll	z=x+y
2	1	2	3
7	2	1	3

P[z=3]= 2/36 = 1/18

P[z=7]=1/6

p[z=5]=1/9

p[z=3]=1/18

(c) probability that at least one 6 appears given that z = 8

Events that favor z=8 are the rows where z=8 from the table are:

#	x:First Roll	y:Second roll	z=x+y
12	2	6	8
17	3	5	8
22	4	4	8
27	5	3	8
32	6	2	8

Number of events that favor [z=8] = 5

Out of these events that favor at least one 6 appears are 2 events as highlighted in the below table

#	x:First Roll	y:Second roll	z=x+y
12	2	6	8
17	3	5	8
22	4	4	8
27	5	3	8
32	6	2	8

Probability that at least one 6 appears given that [z = 8]

= Number events that favor at least one 6 appear and [z=8] / Number of events that favor [z=8]

= 2 / 5

Probability that at least one 6 appears given that [z = 8] = 2/5

(d) probability that x = 6 given that z > 8

Events that favor z> 8 are the rows where z>8 from the table are:

#	x:First Roll	y:Second roll	z=x+y
18	3	6	9
23	4	5	9
24	4	6	10
28	5	4	9
29	5	5	10
30	5	6	11
33	6	3	9
34	6	4	10
35	6	5	11
36	6	6	12

i.e Number of events that favor event z>8 = 10

Out these events, event that favor x=6 are :

#	x:First Roll	y:Second roll	z=x+y
33	6	3	9
34	6	4	10
35	6	5	11
36	6	6	12

i.e Number of events that favor x=6 given z>8 = 4

probability that x = 6 given that z > 8

= Number events that favor x=6 and z>8 / Total number of events that favor z>8 = 4/10 = 2/5

probability that x = 6 given that z > 8 = 2/5

(e) probability that z = 7 given that at least one 4 was rolled

Events that favor at least one 4 is rolled:

#	x:First Roll	y:Second roll	z=x+y
4	1	4	5
10	2	4	6
16	3	4	7
19	4	1	5
20	4	2	6
21	4	3	7
22	4	4	8
23	4	5	9
24	4	6	10
28	5	4	9
34	6	4	10

Number of events that favor at least one 4 rolled =11

From these events , events that favor [z=7] are two

#	x:First Roll	y:Second roll	z=x+y
16	3	4	7
21	4	3	7

probability that [z = 7] given that at least one 4 was rolled

= Number of events that favor [z = 7] and at least one 4 was rolled / Number of events that favor event at least one 4 was rolled = 2/11

probability that [z = 7] given that at least one 4 was rolled = 2/11

(f)The probability that z > 7 given that y = 4.

Events that favor y=4 are the rows from the table where y=4 are as follows

#	x:First Roll	y:Second roll	z=x+y
4	1	4	5
10	2	4	6
16	3	4	7
22	4	4	8
28	5	4	9
34	6	4	10

Number of events favoring [y=4] = 6

Out of these events, events that favor z>7 are

#	x:First Roll	y:Second roll	z=x+y
22	4	4	8
28	5	4	9
34	6	4	10

Number of events that favor z>7 and [y=4] = 3

probability that z > 7 given that y = 4

= Number of events that favor z>7 and [y=4] / Number of event that favor y=4

= 3/6 =1/2

probability that z > 7 given that [y = 4] =1/2