Four couples at a party play a game. They decide to play in teams of two and sel

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 72 74 93 76 05 51 36 51 34 Trial 2 19 76 04 49 97 32 28 61 60 51 03 Trial 3 32 29 63 42 01 08 11 64 67 29 50 Trial 4 50 25 56 20 32 98 04 87 03 99 17 Trial 5 18 47 75 89 96 29 89 24 32 60 39 Based on the simulation, it is about % likely.

Explanation / Answer

In trial 1, we see there is only 1 case out of the 11 cases, where the original couple are tagged together that is 01.

In trial 2, we see there are 2 cases out of the 11 cases, where the original couple are tagged together that is 76, 23

In trial 3, we see there are 3 cases out of the 11 cases, where the original couple are tagged together that is 32,67,01

In trial 4, we see there are 2 cases out of the 11 cases, where the original couple are tagged together that is 32,98

In trial 5, we see there are 3 cases out of the 11 cases, where the original couple are tagged together that is 89,8932

Therefore the average number of cases in total 11*5 trials would be:

= (1+ 2 + 3+ 2 +3)/55 = 0.2

Therefore 20% is the required answer here.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.