So, i decided to bump this thread up. I am back without scanned papers. Just Latex. For the obvious reasons ill just provide only 2 solutions.

I solved it this way:

ans:

Using DeMorgan we get to this point:

After simplication, i am getting here:

But:

We know the rest values from given so i conclude =0.25

My Big Question, given all these facts , is my analysis correct ? The question says that A,B,C are events of a random experiment therefore i think that A,B,C ⊆ Ω , so my analysis should be correct. Any1 to confirm ?

B)

And something else, i am learning combinations as well. So lets say we have this standard example: we have this word --> 'ABCADEFGHEI' 11 Words in total. 2*A and 2*E . How many anagrams of this word can we do provided that the 1st letter is always G and both E's will be next each other.

So, without much thought i came up to this: 9!/2! . But wouldnt an answer like (9!*2!/2!) make more sense? My point is this; the boxed EE's can be arranged in 2! ways (sounds stupid i know, the result is always the same EE or EE). So, 11 letters - 1 (Letter G) =10 Letters . Then, the EE's are boxed so it counts as a single letter, thus we getting a total of 9 letters. And i divide by 2! due to the AA's.

What am i missing?

Thanks a lot.