Four distinct integers are chosen randomly and without replacement from the first twelve positive integers. Let [math]X[/math] be the random variable representing the second largest of the four selected integers, and let [math]p[/math] be the probability function for [math]X[/math].

Determine [math]p(x)[/math], for integer values of [math]x[/math], where [math]p(x)\gt0[/math].

• [math]\frac{( x − 1)( x − 2)(12 − x)}{990}[/math]
• [math]\frac{( x − 1)( x − 2)(12 − x)}{495}[/math]
• [math]\frac{( x − 1)(12 − x)(11 − x)}{495}[/math]
• [math]\frac{( x − 1)(12 − x)(11 − x)}{990}[/math]
• [math]\frac{(10 − x)(12 − x)(11 − x)}{990}[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: A

Because there must be two smaller values and one larger value than X, X cannot be 1, 2, or 12. If X is 3, there is one choice for the two smallest of the four integers and nine choices for the largest integer. If X is 4, there are three choices for the two smallest of the four integers and eight choices for the largest integer. In general, if X = x, there are (x – 1) choose (2) choices for the two smallest integers and 12 – x choices for the largest integer. The total number of ways of choosing 4 integers from 12 integers is 12 choose 4 which is 12!/(4!8!) = 495. So the probability that X = x is:

[[math]] \frac{\binom{x-1}{2}(12-x)}{495} = \frac{(x-1)(x-2)(12-x)}{990}. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.