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12 June, 14:23

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?

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  1. 12 June, 14:25
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    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
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