Question

Four couples at a party play a game. They decide to play in teams of two and select the teams randomly. All eight people write their names on slips of paper. The slips are mixed, then drawn two at a time. How likely is it that every person will be teamed with someone other than who he or she came with? Run five trials using the random digits below. Use the digits 0 and 1 for the first couple, 2 and 3 for the second couple, etc. Trial 1 41 01 56 54 92 57 03 31 27 31 24 Trial 2 28 37 06 68 83 54 49 72 70 12 05 Trial 3 73 39 67 03 12 18 22 60 64 39 51 Trial 4 57 15 52 17 31 89 74 90 73 88 60 Trial 5 18 64 45 89 93 29 89 26 72 30 79 Based on the simulation, it is about 60 % likely.

Answer 1

0/1 , 2/3 , 4/5, 6/7   for each couple

if 8,9 is used it NA

probability that someone will be paired with other than who or she came with

hence for AB if A and B belong to other categories, then it is true

for example

41 is true as 4 and 1 from difference category

41	1
1	0
56	1
54	0
92	NA
57	1
3	1
31	1
27	1
31	1
24	1
28	NA
37	1
6	1
68	NA
83	NA
54	0
49	NA
72	1
70	1
12	1
5	1
73	1
39	NA
67	0
3	1
12	1
18	NA
22	0
60	1
64	1
39	NA
51	1
57	1
15	1
52	1
17	1
31	1
89	NA
74	1
90	NA
73	1
88	NA
60	1
18	NA
64	1
45	0
89	NA
93	1
29	1
89	NA
26	1
72	1
30	1
79	1
	35

required probability= 35 /(55 - 14) = 35/41 = 0.8536 = 85 %