MATH SOLVE

4 months ago

Q:
# Four couples at a dinner party play a board game after the meal. They decide to play as teams of two and to select the teams randomly. All eight people write their names on slips of paper. The slips are thoroughly mixed, then drawn two at a time. How likely is it that every person will be teamed with someone other than the person he or she came to the party with?

Accepted Solution

A:

Answer:0.29Step-by-step explanation:We are given that there are 4 couples i.e. 8 peopleThey decide to play as teams of two and to select the teams randomlyNow we are asked How likely is it that every person will be teamed with someone other than the person he or she came to the party withFor people 1 , there are 6 options for pairing Since he or she cannot pair with his /her own partner So, Choices For people 1 will be 6 personSo, Probability that person 1 will be teamed with someone other than the person he or she came to the party with = [tex]\frac{\text{No. of choices for Person 1}}{\text{Total no. of choices available}}=\frac{6}{7}[/tex]So, Probability that every person (=8 person) will be teamed with someone other than the person he or she came to the party with =[tex](\frac{6}{7})^8=0.29[/tex]Hence Probability that every person will be teamed with someone other than the person he or she came to the party with is 0.29