The answer appears to be three hundred eighty five can't wait to get those points! Hope your happy with your answer!
The answer is <span>4.84166667</span>
Answer:
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
Answer:
The first one is a the second one is b
Step-by-step explanation:
Based on the 5 trials, 1 carton of eggs is expected to contain at least one broken egg, on average
<h3>How to determine the number of cartons</h3>
From the question, we understand that only one of the first 10 double-digit numbers is between 01 and 08
This means that:
The average carton or the expected value of eggs in the simulation is 1
Hence, 1 carton of eggs is expected to contain at least one broken egg, on average
Read more about expected values at:
brainly.com/question/15858152
