May have a few other questions for FG as well if anyone is interested. Thanks!

expected value of X is the sum of the possible values for X multiplied by their probabilities.
In your example, X can take 3 different values :
+1$ (with probability 2/6)
+2$ (with probability 3/6)
+25$ (with probability 1/6)

Expected value, hence, is :

E(X) = (2/6)*1$ + (3/6)*2$ + (1/6)*25$
E(X) = 1/3 + 1 + 25/6
E(X) = 33/6 = 11/2 = 5.5$

If the game costs 5$ to play, then you can expect to make (5.5 - 5) = 0.5$ each time you play it.

Thank you!

I have 2 more I'm having issues with. If anyone could walk me through them I would appreciate it!

5 short problems to choose among 10 : there's C(10,5) choices.
C(10,5) = 10! / (5!*5!) = 10*9*8*7*6 / (5*4*3*2*1) = 252

3 long problems to choose among 8 : C(8,3) choices.
C(8,3) = 56

Different exams that can be made : 252*56 = 14112

C(n,p) denotes the number of different subsets containing p elements in a set containing n elements.

*****

Briefcases game : how many briefcases are black or contain $2,000 ?
There are 6 (5 black and 1 red containg $2,000).
You choose randomly among 9 : there are 6 chances out of 9 to win.
6/9 = 2/3

This post was edited by feanur on Apr 2 2020 12:51am