Eduardo,Kim,Maria,Larry,Ian,Sarah,and Dawn have all been invited to a dinner party. They arrive randomly and each arrives at a different time.
A. In how many ways can they arrive
B. In how many ways can Eduardo arrive first and Dawn last
C.find the probability that Eduardo will arrive first and Dawn last.
A. In how many ways can they arrive?
There are 7 people in total, so there are 7 possibilities for the first person to arrive, 6 possibilities for the second person, and so on. Hence, there are 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040 ways for them to arrive.
B. In how many ways can Eduardo arrive first and Dawn last?
If Eduardo arrives first, there are 6 possibilities for the second person, then 5 possibilities for the third person, and so on, until there are 2 possibilities for the second to last person. Then there is only 1 possibility for Dawn to arrive last. Hence, there are 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 ways for Eduardo to arrive first and Dawn to arrive last.
C. Find the probability that Eduardo will arrive first and Dawn last.
The probability is the number of favorable outcomes divided by the number of possible outcomes. In this case, the number of favorable outcomes is 720 (from part B), and the number of possible outcomes is 5,040 (from part A). Therefore, the probability is 720/5,040 = 1/7 ≈ 0.143.