Answer:
0.22% probability that the six teams chosen were the actual six teams selected to play
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
The order which the teams are chosen is not important, so we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
Desired outcomes:
Six teams chosen were the actual six teams selected to play.
So
[tex]D = C_{6,6} = \frac{6!}{6!(6-6)!} = 1[/tex]
Total outcomes:
Six teams chosen from a set of 11. So
[tex]T = C_{11,6} = \frac{11!}{6!(11-6)!} = 462[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{1}{462} = 0.0022[/tex]
0.22% probability that the six teams chosen were the actual six teams selected to play
