Three things are needed to determine the expected value of a game - the probability of a win, the value of a win, and the value of a loss.

Statement 1 provides insufficient information. The statement does not give the number of tickets sold, so no clue as to the probability of a win is given. Also, neither the value of a win not that of a loss is given; the worth of the prize is given, but without the price of the ticket, the value of the win (the worth of the prize minus the price of a ticket) cannot be determined.

Statement 2 provides insufficient information. The statement gives the value of a loss: , the cost of a ticket. The number of tickets sold is given to be 400, so the probability of a win is  and that of a loss is ; however, the statement does not give the worth of the prize, which is needed to find the value of the win.

Assume both statements hold. The value of a win is $795—the $800 value of the laptop minus the $5 ticket price. The value of a loss is , as follows from Statement 2. The probabilities are those determined from Statement 2. The expected value of a ticket is the sum of the products of the probability and vlue of each outcome:

which justifies Ted purchasing a ticket under his self-imposed criterion.

(1) A study conducted in the hospital revealed that   of the patients will have at least two tests performed.

Using the information in Statement (1), we cannot find the probability of patients who have exactly two tests performed upon entering the hospital. The probability given in Statement (1) also includes patients who have more than 2 tests performed.

Therefore, Statement (1) Alone is not sufficient.

(2) A study conducted in the hospital revealed that 1 in 5 patients will have 2 tests performed.

Using Statement (2), we can calculate the probability that a patient entering the hospital will have exactly two tests performed as:

(1) There are only 3 questions on the quiz.

Using Statement (1), we know that the probability that the student guesses the correct answer to each question is the product of the probabilities of getting each individual question right (getting each question right constitutes independent events). That is:

Pr(Question1 Correct) x Pr(Question2 Correct) x Pr(Question3 Correct)

However Statement (1) Alone is not sufficient.

(2) Each question has 4 answer choices and only one choice is correct.

Using Statement (2), we know that the student has the same probability of getting each question right, which we calculate as:

Pr(Correct Answer)=

Still Statement (2) Alone is not Correct.

Combining Both statements, we find the probability of guessing each correct answer on the quiz as:

Therefore Both statements Together are sufficient but neither Statement Alone is sufficient.

To find probability, we need to know the number of each color of marker. Statement I gives us a way to put the number of red markers in terms of green markers. Statement II tells us the number of yellow markers. Using Statement II, we can find out how many red and green markers we have together. Then we would have to use Statement I to find the number of red markers and green markers. Then we can calculate probability. There is no way to do it without both statements.

To recap:

Bronson has a box of 16 markers. The markers are green, red and yellow.

I) The number of green markers is twice the number of red markers

II) There are 4 yellow markers

What are the odds of pulling a yellow marker followed by a green marker followed by a red marker? (Assume no replacement.)

So:

So, we have 4 yellow markers, 4 red markers and 8 yellow markers.

To find the odds of three events, we need to multiply the probability of each event together. We assume no replacement, so for each event, our total number of outcomes decreases by one.

So, the odds are 4 in 105.

According to the probability table, only an odd number can be rolled. Three of our choices can be eliminated immediately for this reason. 

If the dice have  on their faces, it is possible to roll a double 6 for a 12; if the dice have  or  on their faces, it it possible to roll a double 1 for a 2.

This leaves two possibilites. All we need to look at is the probability of rolling a 3 with each choice.

:

A 3 can only be rolled by rolling a 1 on the first die and a 2 on the second. The probability of doing this is 

:

Again, a 3 can only be rolled by rolling a 1 on the first die and a 2 on the second. The probability of doing this is 

.

The other probabilities can be confirmed to agree with the ones given in the conditions in the problem. This is the correct choice.

Janice has a cooler with bottles of water, cola, and root beer. What are the odds of choosing a water bottle followed by a bottle of root beer?

I) There are 5 times more bottles of root beer than water.

II) There are 75 beverages total in the cooler.

To calculate probabilty, we need to know the number of desired outcomes and the total number of outcomes. 

II) gives us the number of total possible outcomes.

I) gives us a clue about the number of desired outcomes, but it is insufficient  to find the actual number of root beers and waters.

Therefore 

Neither statement is sufficient to answer the question. More information is needed.

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).

Thus, the probability of three heads in five flips is:

The probability of either of two events happening, the union, is given by the formula:

Because we are not told that drawing a blue and drawing a nine are mutually exclusive events, we must know the probability of the intersection, of drawing a blue nine.

Since this probability question deals with successes and failures, it regards binomial distributions:

Where   is the number of trials,  is the number of 'successes,'  is the probability of exactly  successes in  trials, and  is the probability of a success occuring.

Statement 1 allows for a calculation of  to be used in the probability calculation for the problem.

Statement 2 outright gives a  to calculate the probability asked of in the problem.

Since this probability question deals with successes and failures, it regards binomial distributions:

Where   is the number of trials,  is the number of 'successes,'  is the probability of exactly  successes in  trials, and  is the probability of a success occuring.

The probability of at least one success is the complement of exactly zero successes. Since we're looking at 3 trials, we can find the probability of zero successes to be  and thus the complement, our .

For fewer than three trials, the probability will always be below , but for three or more, it depends on exactly how many trials there are, since the probability increases with increasing number of trials.