the subway train has 18 doors and starts with 15 passenger. if each passenger is equally likely to get off at any station and the passengers leave the train independently, find the probability that 2 or more passengers leave the train through the same door
To find the probability that 2 or more passengers leave the train through the same door, we can use the concept of the complement rule.
First, let's find the total number of ways in which the 15 passengers can leave the train through 18 doors. This can be calculated using combinations. Since each passenger is equally likely to get off at any station and the passengers leave independently, we can find the total number of ways as:
Total number of ways = number of ways to choose 15 doors out of 18 doors = C(18,15) = 18! / (15! * (18-15)!) = 816
Now, let's find the number of ways in which no two passengers leave through the same door.
The first passenger can choose any of the 18 doors, the second passenger can choose any of the remaining 17 doors, the third passenger can choose any of the remaining 16 doors, and so on. Therefore, the number of ways in which no two passengers leave through the same door is:
Number of ways = 18 * 17 * 16 * ... * 4 * 3 * 2 * 1 = 18!
Now, let's find the probability that no two passengers leave through the same door:
P(no two passengers leave through the same door) = Number of ways / Total number of ways = 18! / 816
Finally, we can find the probability that 2 or more passengers leave through the same door by using the complement rule:
P(2 or more passengers leave through the same door) = 1 - P(no two passengers leave through the same door) = 1 - (18! / 816)
Evaluating this expression will give us the final probability.