Half of the scrunchies are black, so

One third of the scrunchies are white

One ninth of the scrunchies are blue

And the rest are green:

If we have already drawn two, our total amount of scrunchies is now 88, so the probability of pulling a black scrunchie will be:

of the students are wearing red shirts, and

One third of the students are wearing pink shirts

The remaining students are wearing orange

The probability of randomly selecting a student wearing an orange shirt is

There are four yellow candies, and 12 orange candies, so the raio is:

The ratio of yellow to orange candies is 1:3

Recall:

Consequently, the probability that the person selected is Asian is:

Similarly, the probability that a randomly selected Asian person is also female is:

Finally, the probability of 2 unrelated events occuring is equal to the product of the individual probabilities of the 2 events. Therefore the probability of selecting an Asian female is

Fred can wear anything to school. This means he can wear either formal things or informal things.  He has 3 informal shirts and 2 formal shirts.  This means he has a total of 5 shirts to choose from when deciding what to wear.  Similarly, he has 3 pairs of pants and 2 pairs of shoes to choose from.  After choosing a shirt (5 options), he chooses pants (3 options) and shoes (2 options) resulting in  choices for outfits to wear to school.

Fred must wear formal attire to the dinner.  This means that Fred can only choose between his 2 formal shirts, his formal pair of pants, and his formal shoes.  Since he only has a choice of shirts, he ends up only having 2 choices to make in outfits.

To figure out how many ways he can dress throughout the day, we simply choose an outfit for school and an outfit for dinner.  We have 30 choices for what to wear to school and 2 choices to wear to dinner, giving us choices in total.

Firstly, when taking a percentage, we can simply multiply by the proper decimal, where multiplying by 1 is 100%.

The school has 2,000 students.  Since only 85% responded, we have  students that were not accounted for. 

Of the 1700 who responded,  are male.  Of these students,  believe that there should be more invested in extracurriculars.

The key is recognizing that we can 'control' the 300 votes that are unaccounted for.  These students were not counted, but might be male or female, and might vote for or against increased spending on extracurriculars.

Thus while we initially have 612 males believing in extracurriculars, it might be true that the remaining 300 uncounted votes were all males that believed in increasing spending on extracurriculars, resulting in up to 912 males voting for it.

All the other options are entirely impossible. 

We know that  are in favor of spending on extracurriculars.  This means that the majority of students believe in increased spending on extracurriculars not on core classes, and that more than  of the school believe that this spending should be increased.

We also know that, even should we have the maximum possible of 912 males in favor of increasing spending on extracurriculars, the 935 females who are in favor are still greater.

Finally, the number of male students who prefer extracurriculars is currently 612.  Even should all 300 remaining voters be males in favor of core classes, we end up with  in favor of core classes vs 612 in favor of extracurriculars.

If the first toss is a 1, it can not be bigger than any of the numbers on the second toss. If first toss is a 2, then it can be bigger than 1. If the first toss is 3, it can be bigger than 1 or 2. With this pattern, if the first toss is a 6, it can be bigger than any of the 5 numbers on the second toss.

Therefore, we get  ways to have the first toss being bigger than the second toss. The total number of combinations is .

So we have a probablility of .

There are four combinations total, being: {(odd,odd), (odd, even), (even, odd), (even, even)}

Given that their product is even, we only have the set {(odd,even),(even,odd),(even, even)}

The probability that their sum is odd from the set {(odd,even),(even,odd),(even, even)} is  because  and .

The first die comes up as a 5.  We are essentially asking about outcomes of two dice, plus 5.  As a first step, we should be sure we know the primes we need to exempt.  The smallest sum we can have is if both of the other dice roll 1, giving us a 7.  7 is a prime number, so will will exempt this number. The largest number we can roll is both rolling 6, giving us a total of 17 (also a prime).  All the primes we will need to remove are then: 7,11,13 and 17.

We have two approaches now.  We can find the odds of rolling a number besides those above directly, or we can take the odds of getting any of the above 4, and then subtracting this value from 1.

The latter method is faster, but both use the same method.  Firstly, we'll subtract 5 from all the numbers: 2,6,8,12. 

The odds of getting a 2 on two dice is , since the only roll that will work is two 1s and there are a total of 36 possible rolls (6 choices per die). 

Similarly, the odds of getting a 12 is  since it requires two 6's.

For 6, you can roll 1-5, 2-4, 3-3, 4-2, 5-1.  This gives us  odds of getting a 6.  Similarly, an 8 can be achieved by rolling 6-2, 5-3, 4-4, 3-5, 2-6.

Our total odds of rolling any of these 4 is thus:

Our answer is then: