Question

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?

Answer 1

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