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Thread: [SOLVED] cinema problem

  1. #1
    Junior Member JoanF's Avatar
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    [SOLVED] cinema problem

    4 friends go to the cinema and reserve 4 tickets. The seats are already assigned to them. When they arrive to the cinema, they sit at random. The probability of 1 stay in its place is greater or minor than or equal to the probability of 2 sit in its place?
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  2. #2
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    Hello, JoanF!

    4 friends go to the cinema and reserve 4 tickets.
    The seats are already assigned to them.
    When they arrive to the cinema, they sit at random.

    The probability that one is in the right seat is:
    . . (1) greater than
    . . (2) less than
    . . (3) equal to
    the probability that two are in the right seats?
    There are $\displaystyle 4! = 24$ possible seatings.

    Suppose the four friends are: .$\displaystyle \{A,B,C,D\}$
    . . and their assigned seats are in this order: .$\displaystyle A\:B\:C\:D$


    One in the right seat.

    There are 4 choices for the One.
    . . Suppose it is $\displaystyle A.$

    Then there are 2 ways for the others to sit in the wrong seats:
    . . $\displaystyle A\:C\:D\:B\;\;\text{ and }\;\;A\:D\:B\:C$

    Hence, there are: .$\displaystyle 4 \times 2 \:=\:8$ ways.

    . . $\displaystyle P(\text{one in the right seat}) \:=\:\frac{8}{24}\:=\:\frac{1}{3}$


    Two in the right seats.

    There are: .$\displaystyle {4\choose2} = {\color{blue}6}$ choices for the Two.
    . . Suppose they are $\displaystyle A$ and $\displaystyle B.$

    Then there is 1 way for the other two to be in the wrong seats:
    . . $\displaystyle A\:B\:D\:C$

    Hence, there are: .$\displaystyle 6 \times 1 \:=\:6$ ways.

    . . $\displaystyle P(\text{two in the right seats}) \:=\:\frac{6}{24}\:=\:\frac{1}{4}$


    Therefore: .$\displaystyle P(\text{one in right seat}) \;\;{\color{red}>}\;\;P(\text{two in the right seats}) $

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