Problem 1) No the events are not mutually exclusive. Any time you roll doubles, the sum will always be even. Let x be any integer between 1 and 6. If we roll two 'x's, then we have x+x = 2x as the sum. By definition 2x = 2*x is an even integer.
A specific example is if we rolled a pair of '3's an we get 3+3 = 6 as a result.
So in conclusion, the events "rolling doubles" and "getting an even sum" are not mutually exclusive. If the first event happens, then the second event definitely happens. The same can't be said the other way around.
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Problem 2)
We have 8 students. Of those 8, we have two students Laura and Kimiko that we want to keep together. In other words, we want them to be adjacent to each other. If we think of Laura and Kimiko as one student, then we have 8-2+1 = 7 "students", 6 of which are individuals who can move on their own and 2 (Laura and Kimiko) who are tied together as one "student".
There are 7! = 7*6*5*4*3*2*1 = 5040 ways to arrange these 7 "students". For any given permutation, we can have Laura on the left side or Kimiko on the left side. So there are 2 ways to arrange Laura and Kimiko.
In total there are 2*5040 = 10080 different ways to arrange all of the students where Laura and Kimiko will be next to each other.
This is out of 8! = 8*7*6*5*4*3*2*1 = 40320 ways to arrange the students where we don't worry about Laura and Kimiko sticking together.
Divide the two values: 10080/40320 = 1/4
Answer:
The answer as a fraction is 1/4
The answer as a decimal value is 0.25 (since 1/4 = 0.25)
In percent form, the answer is 25% (0.25 turns into 25% after moving the decimal 2 spots to the right)