### All GRE Math Resources

## Example Questions

### Example Question #61 : Data Analysis

A card is drawn at random from a deck of fifty-two cards (no jokers). What is the probability of drawing a diamond, a card with an even number, or a card with a number divisible by three?

**Possible Answers:**

**Correct answer:**

First, consider the probability of each individual event: drawing a diamond, an even number, or a number divisible by three.

Drawing a diamond, there are four suits, so:

Drawing an even number, there are five possible values present in each thirteen-card suit, so:

Drawing a number divisible by three, there are three possible values present in each thirteen-card suit, so

This problem deals with a union of probabilities, essentially a "this or that" option; however, since there are three events considered, the formula for the union follows the form:

The probability of drawing an even number that is also a diamond is:

The probability of drawing a diamond that is divisible by three is:

The probability of drawing an even number that is divisible by three is:

And the probability of drawing a number that is a diamond, even, and divisble by three is:

Therefore, the probability of this union is:

### Example Question #54 : How To Find The Probability Of An Outcome

Alex and Jeffery both apply to the same college that randomly accepts students. If Alex has a chance of being accepted and Jeffery has a chance of being accepted, what is the probability of the college accepting Jeffery and not Alex?

**Possible Answers:**

**Correct answer:**

To solve this problem, we must first understand a basic concept of probability. In order to find the probability of two independent events happening, we multiply the probability of each of those two independent events happening together in order to find the probability of both of them occurring.

In this case, we know that Alex has a chance of being accepted and chance of being rejected.

We also know that Jeffery has a chance of being accepted and a chance of being rejected.

The question asks us the probability of the college accepting Jeffery, but not Alex, therefore we multiply the probability of the college accepting Jeffery, , by the probability of the college rejecting Alex, , together in order to answer this question.

or a chance of Alex being rejected by Jeffery being accepted.

### Example Question #55 : How To Find The Probability Of An Outcome

If the probability of Martin losing his car keys next month is , that is the probability that Martin will not lose his car keys next month?

**Possible Answers:**

**Correct answer:**

To solve this problem, we must first understand some of the basic concepts of probability. If the probability of an outcome is , that means that there is a chance of that outcome happening. Thus the probability of an outcome occurring can never exceed

This question asks the probability of Martin NOT losing his car keys. It tells us that the probability of him losing his car keys is , or . Because we know there is a chance he will either lose or keep his car keys and we know that probability of him losing his car keys, we simply subtract the probability of him losing his car keys from the total chance, or 90.25% chance that Martin will not lose his car keys next month.

### Example Question #56 : How To Find The Probability Of An Outcome

A candy jar has skittles and M&M's. What is the percent chance, when randomly selecting two candies in a row without replacement, that both will be skittles? Round to the nearest percent.

**Possible Answers:**

**Correct answer:**

To solve this problem, we must understand some basic concepts. In the jar, there are currently candies that can be chosen. Of these are skittles and are M&M's. This means that there is a or chance when randomly selecting a candy for the first time that it will be a skittle.

The question asks us the probability of selecting two skittles in a row, therefore we must once again calculate the probability of selecting a skittle a second time. After picking a skittle the first time, there are now candies in the jar and only of them are skittles now. This means that there is a chance of randomly selecting a skittle the second time.

Now that we have our two probabilities, to find the probability of both happening, we simply multiply them together, meaning there is chance of approximately chance of grabbing two skittles in a row from the jar.

### Example Question #57 : How To Find The Probability Of An Outcome

Terry draws cards from a standard card deck. What is the probability of him drawing a face card first and a club next?

**Possible Answers:**

**Correct answer:**

There are twelve face cards (king, queen, and jack for each of the four suits) in a deck, so the probability of drawing a face card on the first draw is

Now, there are two potential situations for the second draw given he draws a face the first time; either he drew a face from the clubs or he didn't.

The probability of a face also being a club is

And the probability of the complement is then:

Were a face club drawn, the probability of drawing a club next would be

Since a club was removed from the deck.

Were a face club not drawn, the probability of drawing a club next would be

Therefore, the unconditional probability of drawing a club accounts for the probability of either of these scenaorios:

Now, for the probability of drawing the face first and the club next, it is the intersection of events:

### Example Question #58 : How To Find The Probability Of An Outcome

Salisbury is at an ice cream shop that is somewhat limited in selection, but with a quality that makes up for it. There are ten flavors of ice cream and four kinds of toppings. If Salisbury orders the two-pair special, which is two random unique ice cream scoops and two random toppings, what is the probability of him getting vanilla and chocolate with almonds and sprinkles?

**Possible Answers:**

**Correct answer:**

In this problem, the order of selection for ice cream and toppings does not matter, so it's dealing with combinations.

The number of possible combinations for selections out of items is:

So the number of possible ice cream choices in this problem is:

And the chance of choosing a chocolate+vanilla selection is .

For the toppings, the number of combinations is:

With the almond sprinkle combination having a chance of being chosen.

The chance of BOTH of these selections being made is then the product of the two probalities:

### Example Question #59 : How To Find The Probability Of An Outcome

At a specialty restaurant, there are five kinds of curry and three kinds of rice. If a lunch special consists of two unique curries and a side of rice, what is the total number of possible options?

**Possible Answers:**

**Correct answer:**

For this problem, when selecting food items, the order of selection does not matter, so we're dealing with combinations.

For choices made from possible options, the number of possible combinations is

So for the curry options, where two choices are made from five options

For the rice options, with one choice made from three:

For the lunch special, the total amount of combinations is the product of these two:

### Example Question #60 : How To Find The Probability Of An Outcome

A twenty-sided die is rolled three times. What is the probability of rolling a five or greater all three times?

**Possible Answers:**

**Correct answer:**

For a twenty-sided die, there are twenty possible rolls:

Of these, sixteen satisfy the problem condition:

So the probability of rolling a five or greater is

The probability of rolling three of these rolls in a row is a type of intersection, meaning multiplication is involved:

### Example Question #61 : How To Find The Probability Of An Outcome

A twenty-sided die is rolled three times. What is the probability of rolling a five or greater at least twice?

**Possible Answers:**

**Correct answer:**

For a twenty-sided die, there are twenty possible rolls:

Of these, sixteen satisfy the problem condition:

So the probability of rolling a five or greater is

Since there's a probability of failure or success (rolling more than five, or not rolling more than five), this is a binomial distribution.

The probability of exactly successes out of trials, where a success has a probability , for a binomial distribution is

Since we're considering the probability of at least two successes out of three, we must consider the two conditions that satisfy this: two rolls that are greater than five, or three rolls that are greater than five. The total probability will be the sum of these two probabilities.

The total probability is

### Example Question #61 : Probability

A coin is flipped ten times.

Quantity A: The probability of getting heads at least nine times.

Quantity B: The probability of getting heads eight times.

**Possible Answers:**

The relationship cannot be determined.

Quantity A is greater.

The two quantities are equal.

Quantity B is greater.

**Correct answer:**

Quantity B is greater.

This problem is dealing with the repeated trials for an event that only has two potential outcomes. This scenario describes a binomial distribution.

For trials, each of which has a probability of a 'successful' outcome, the probability of exactly successes is given by the function:

For Quantity A:

We're considering the probability of getting at least nine heads out of ten flips. This means either flipping head nine times OR ten times.

The probability is

For Quantity B:

We're considering the probability of exactly eight flips out of ten.

Quantity B is greater.