Here are 100 seats on an airplane, and each of 100 passengers has a ticket for a different seat. the passengers line up to board the plane, but just before boarding, the first passenger loses his ticket. since he has no idea where to sit, he chooses aseat uniformly at random from the 100 seats. each subsequent passenger first goes to their assignedseat. if that seat is empty, then they sit down, but if it is occupied then they choose a seat at randomfrom the remaining empty seats. this proceeds until the last person boards the plane. what is the probability that the last person gets to sit in their assigned seat?

There are 1% probability that the last person gets to sit in their assigned seat

Step-by-step explanation:

The probability that the last person gets to sit in their assigned seat, is the same that the probability that not one sit in this seat.

If we use the Combinatorics theory, we know that are 100! possibilities to order the first 99 passenger in the 100 seats.

LIke we one the probability that not one sit in one of the seats, we need the fraction from the total number of possible combinations, of combination that exclude the assigned seat of the last passenger. In other words the amount of combination of 99 passengers in 99 seats: 99!

Now this number of combination of the 99 passenger in the 99 sets, divide for the total number of combination in the 100 setas, is the probability that not one sit in the assigned seat of the last passenger.

P = 99!/100! = 99!/ (100 * 99!) = 1/100

There are 1% probability that the last person gets to sit in their assigned seat

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2

9 sandwiches per hour

step-by-step explanation:

when we have a rate, we want a 1 in the denominator

27/3 = 9

27 sandwiches/ 3 hours = 9 sandwiches/hour