Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

person gives n=7 gifts to 6 friends. friends are picked randomly. Find the probability that at least 1 friend does not receive a gift.?

1 Answer

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  • 1 month ago

    Let's first count the total way that we can distribute the gifts.

    There are 6 choices of recipient for each of the 7 gifts (repeats allowed obviously).

    6^7 = 279,936 ways

    Now rather than looking for the ways to hand out the gifts so that at least one person *doesn't* get a gift, let's count the ways that everyone can get at least one gift.

    If we hand out the gifts, one person will get 2 gifts and the rest will get 1.

    First pick the person that gets the two gifts:

    6C1 = 6 ways

    Then pick the 2 gifts that person gets:

    7C2 = (7 * 6) / 2 = 21 ways

    Finally there are 5! = 120 ways to hand out the remaining gifts to the other 5 friends.

    6 * 21 * 120 = 15,120 ways to hand out gifts so that EVERYONE has at least one gift.

    Subtract and you see that there are 264,816 ways to hand out the gifts so that some people don't get gifts.

    Now divide that by the total ways to distribute gifts (279,936) to get the probability.

    Answer:

    264816/279936

    ≈ 0.94598765432

    About 94.6%

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