Haroldo, Xerxes, Regina, Shaindel, Murray, Norah, Stav, Zeke, Cam, and Georgia are invited to a dinner party. They arrive in a random order and all arrive at different times. What is the probability that Xeres arrives first AND Regina arrives last?

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:

In this problem:

Hence, the probability is:

0.0111 = 1.11% probability that Xeres arrives first AND Regina arrives last.

A similar problem is given at link

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

its 4 because 2 times 2 is 4, if you double the number two you get four

or