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.

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:

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 rolling a four), and  is the probability of success (one in four).

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).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

Therefore, the probability of six or fewer heads is:

Set A: 

Set B: 

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is .

The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:

Set A: 

Set B: 

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is .

The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:

The interesection is:

So, we can find the probability of drawing at least one consonant: