After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. at the beginning of the next lecture, the professor distributes the four books in a completely random fashion to each of the four students (1, 2, 3, and 4) who claim to have left books. one possible outcome is that 1 receives 2’s book, 2 receives 4’s book, 3 receives his or her own book, and 4 receives 1’s book. this outcome can be abbreviated as (2, 4, 3, 1). a. list the other 23 possible outcomes. b. let x denote the number of students who receive their own book. determine the pmf of x.

List of outcomes:

{1, 2, 3, 4}, {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {1, 4, 2, 3}, {1, 4, 3, 2},

{2, 1, 3, 4}, {2, 1, 4, 3}, {2, 3, 1, 4}, {2, 3, 4, 1}, {2, 4, 1, 3}, {2, 4, 3, 1},

{3, 1, 2, 4}, {3, 1, 4, 2}, {3, 2, 1, 4}, {3, 2, 4, 1}, {3, 4, 1, 2}, {3, 4, 2, 1},

{4, 1, 2, 3}, {4, 1, 3, 2}, {4, 2, 1, 3}, {4, 2, 3, 1}, {4, 3, 1, 2}, {4, 3, 2, 1}

pmf:

  0: 9/24, 1: 8/24, 2: 6/24, 3: 0, 4: 1/24

Step-by-step explanation:

It works out nicely to let a math program list the permutations and count the numbers that are in the right place. (See below.) This can be done by hand, but is tedious and prone to error.

(Note that it is not possible for exactly 3 students to receive the correct book.)

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step-by-step explanation: