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:

The penny will come up heads with probability 0.54 and tails with probability 0.46. The nickel will come up heads with probability 0.46 and probability 0.54. 

The outcomes of the tosses of the penny and the nickel are independent, so the probabilities can be multiplied. The probability of the penny being heads and the nickel being tails is . The probability of the reverse happening is .

These are disjoint, so add these probabilities; the probability that one head and one tail will come up will be , or 50.32 %. This rounds to 50%, which is not among the choices.

6 to 5 odds in favor of heads is equal to a probability of , which is the probability that the penny will come up heads. The probability that the penny will come up tails is .

Similarly, 3 to 2 odds in favor of heads is equal to a probability of , which is the probability that the nickel will come up heads. The probablity that the nickel will come up tails is .

The outcomes of the tosses of the penny and the nickel are independent, so the probabilities can be multiplied.

The probability of the penny and the nickel coming up heads is .

The probability of the penny and the nickel coming up tails is .

The probabilities are added:

, making the odds 28 to 27 in favor of both coins coming up the same.

Let  represent the number of red marbles in the bag,  represent the number of blue marbles, and  represent the number of green.

We know the total number of marbles is 78:

We know there are 2 times as many red marbles as blue:

And we know there are 4 times as many blue as green:

Using substitution, we can see there are 8 times as many red marbles as green:

Now, rewrite the original equation in terms of :

Solve for .

There are 6 green marbles in the bag. Using this number, calculate the number of blue marbles in the bag:

There are 24 blue marbles in the bag. In order to find the probability of choosing a blue marble on the first pick, take the number of blue marbles in the bag and set it against the total number of marbles:

This reduces to:

The probability of choosing a blue marble on the first pick is .

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.

First, find the number of possible three digit numbers that can be created from the lottery machine. Because the order that the numbers comes out matters (since the number 345 is different than 543), you should use the permutations formula:

Because you are choosing 3 balls out of 10, n=10 and k=3 so

This reduces to  which equals 

You can then cancel out most of the numbers, leaving only 

Since you have only one three digit number on your lottery ticket, your probability of having the winning number is 

.

Note: Because the balls are not replaced, this question is really asking "How many 3-digit numbers are there in which no integer appears more than once?" 

“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.