Respuesta :
Answer: The probability is P = 0.056
Step-by-step explanation:
The invited ones are:
Haroldo, Xerxes, Regina, Shaindel, Murray, Norah, Stav, Zeke, Cam, and Georgia
So we have a total of 10 persons.
The case where Xerxes arrives first and Regina arrives last is:
We have 10 slots, each slot corresponds to who arrived in which time.
For the first slot we have only one option, this is Xeres.
For the next slot we have 8 options (because Xeres already arrived, and Regina must arrive at last)
For the next slot we have 7 options.
for the next one we have 6 options, and so on.
So the combinations are:
C = 1*8*7*6*5*4*3*2*1*1 = 40320
Now, the total number of combinations is:
for the first slot we have 10 options, for the second we have 9 options and so on.
c = 10*9*8*7*6*5*4*3*2*1 = 725760
The probability is the combinations where Xerxes arrives first and Regina rrives last divided the total number of probabilities.
P = C/c = 0.056
Using the concept of probability and the arrangements formula, it is found that 0.0111 = 1.11% probability that Xeres arrives first AND Regina arrives last.
A probability is the number of desired outcomes divided by the number of total outcomes.
The number of possible arrangements of n elements is given by:
[tex]A_n = n![/tex]
In this problem:
- 10 people are invited, so the number of ways they can arrive is [tex]T = 10![/tex].
- Xeres first and Regina last, for the middle 8 there are [tex]D = 8![/tex] ways.
Hence, the probability is:
[tex]p = \frac{D}{T} = \frac{8!}{10!} = 0.0111[/tex]
0.0111 = 1.11% probability that Xeres arrives first AND Regina arrives last.
A similar problem is given at https://brainly.com/question/24648661
