Kim,Sarah,Tyrone,Dawn,Simone,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 Kim arrive first and Jim last
c. Find the probability that Kim will arrive first and Jim last.
a. There are 6 people arriving at the dinner party. The number of ways they can arrive is 6 factorial (6!) or 6 x 5 x 4 x 3 x 2 x 1 = 720 ways.
b. If Kim arrives first and Jim arrives last, we need to consider them as fixed positions. This means that there are 4 remaining people (Sarah, Tyrone, Dawn, and Simone) who can arrive in the middle.
The number of ways these 4 people can be arranged is 4 factorial (4!) or 4 x 3 x 2 x 1 = 24 ways.
c. The probability that Kim arrives first and Jim arrives last is the number of favorable outcomes (the number of ways Kim can arrive first and Jim can arrive last) divided by the total number of outcomes (the total number of ways they can arrive).
The number of favorable outcomes is 24 (as calculated in part b).
The total number of outcomes is 720 (as calculated in part a).
Therefore, the probability is 24/720 = 1/30 or approximately 0.0333 or 3.33%.