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:

2:30 = 150 seconds

3:20 = 200 seconds

3:45 = 225 seconds

Here we have a logic statement:

If A (Jean's brother), then B (red hair).

"If Ron has red hair, then he is Jean's brother" states "If B, then A" - we do not know whether or not this is true.

"If Winston does not have red hair, then he is not Jean's brother" states "If not B, then not A" - this has to be true.

"If Paul is not Jean's brother, then he has red hair" states "If not A, then B." We do not know whether or not this is true.

"If Eddie is Jean's brother, then he does not have red hair" states "If A, then not B" - we know this cannot be true.

The probability that a randomly selected customer is in the moderate- or high-risk class is simply the sum of the number of clients in the moderate-risk class and the number of clients in the high-risk class divided by the  total number of clients:

This is also equivalent to just summing up the probability of being in the high-risk class and the probability of being in the moderate-risk class. Since the 3 classes are mutually exclusive, we do not have to worry about subtracting the probability of mutual elements.

“Ben only goes to the park when it is sunny.”  This means it has to be sunny out for Ben to go to the park, but it does not mean that he always is at the park when it is sunny.  Looking at the first choice we can say that this is not necessarily true because it could be sunny and Ben doesn’t have to be at the park.  Similar reasoning would prove the second choice wrong as well.  The third choice is correct-Ben only is at the park when it is sunny, so he’s definitely not there when it’s not sunny.  The fourth choice is clearly wrong because we know Ben only goes when it is sunny, not when it’s rainy.

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is .

There are thirteen spades in the deck, so the probability of drawing a spade is .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is 

Thus the probability of drawing a face card or a spade is:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where  is the number of events,  is the number of "successes" (in this case, a "heads" outcome), and  is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

Thus the probability of three or more flips is: