Question

Dawn comma Sergio comma Tyrone comma and Jim have all been invited to a dinner party. They arrive randomly and each person arrives at a different time. a. In how many ways can they​ arrive? b. In how many ways can Dawn arrive first and Jim ​last? c. Find the probability that Dawn will arrive first and Jim last.

Answer 1

Answer:

a. 24

b. 2

c. 0.0833 = 8.33%

Step-by-step explanation:

a.

The first "slot" of person to arrive has 4 possibilities, then the second "slot" will have 3 possibilities, as one has already arrived, then the third "slot" has 2 possibilities, and the fourth "slot" has just 1 possibility.

So, multiplying all these combinations, we have 4*3*2*1 = 24 possible ways they can arrive

b.

If the first and the last person are already "locked", we just have possibilities for the second and third person. The second will have 2 possibilities (Sergio or Tyrone), and the third will have only 1 (the person that wasn't the second between Sergio and Tyrone). So, the number of possibilities is 2*1 = 2

c.

If we have 2 cases where Dawn is first and Jim is last, from a total of 24 possible cases, the probability is 2/24 = 1/12 = 0.0833 = 8.33%