CPhill

a)  Theoretical probability  =  (1/6)100  = about 16 2/3 %  = about 16 times in 100

b) Experimental probability  = 47/100  = 47%

c) The die does not appear to be "fair"......we would expect a "6"  to occur far less frequently in 100 rolls....however.....although we should expect that each number would appear about 16 times in 100 rolls, this, in fact, will probably not happen.....

Actually, because of  the Earth's curvature - and other factors - the procedure is a little more complicated. 

You can read about it here :  

http://www.pveducation.org/pvcdrom/properties-of-sunlight/elevation-angle

18 have been selected.....9 have  been orange.....so we expect that the probability that the next marble is orange  is 9/18  = 1/2

a)  a combination........combinations count sets......permutations count orderings of sets......here, the order does not matter

b)  C(8,3)  = 56  ways   ....thus 56 different sets

c)  P(3 male)  =  (sets of all possible men) /(all possible sets )  = 

  C(5,3) / C(8,3)  =  10 / 56  =  5 / 28   ≈  17.86 %

sin^2θ-csc^2θ=tan^2θ-cot^2θ

This is not an identity

We can see this from the graph, here....

https://www.desmos.com/calculator/lkwcraopqf

The graph of the left side  is not the same as the graph of the right side

If12 were sold and 3  were green.....then  P(green)  =  3 / 12  =  1 / 4

The day of the week  is given by

(1000 mod 7)  -  1  =  

5 ( days  after Wednesday)  =  Monday

LOL!!!...I'll take their word for it......!!!!

If each successive bounce gets smaller by a factor of 1/3, it will bounce 2/3 of its previous height on each successive bounce......

The third drop will be from a height of 300 (2/3)^2   ≈  133.33 m

Here's the  fourth one 

We are just looking for all the possible arrangememts of this set  (2, 2, 2, 4, 6)

Note  that the "6"  can occupy  any one of 5 positions  and once its placement is determined, the "4" can placed in any  of the 4  other positions.....

So..... the total number of sequences  =  5 * 4  =  20