### All GRE Math Resources

## Example Questions

### Example Question #1 : Probability

A jar contains 10 red marbles, 4 white marbles, and 2 blue marbles. Two are drawn in sequence, not replacing after each draw.

Quantity A** **

The probability of drawing two red marbles

Quantity B

The probability of drawing exactly one blue marble.

**Possible Answers:**

Quantity A is greater.

The relationship cannot be determined from the information given.

Quantity B is greater.

The quantities are equal.

**Correct answer:**

Quantity A is greater.

Note that there are 16 total marbles. A is simply a set of sequential events. On the first, you have 10/16 chances to draw a red. Supposing this red is not replaced, the chance of drawing a second red will be 9/15; therefore, the probability of A is (10/16) * (9/15) = 0.375. Event B is translated into 2 events: Blue + (White or Red) or (White or Red) + Blue. The probabilities of each of these events, added together would be (2/16) * (14/15) + (14/16) * (2/15) = 0.2333333333; therefore, A is more probable.

### Example Question #1 : How To Find The Probability Of An Outcome

In a bowl containing 10 marbles, 5 are blue and 5 are pink. If 2 marbles are picked randomly, what is the probability that the 2 marbles will not both be pink?

**Possible Answers:**

7/9

7/8

2/9

5/6

**Correct answer:**

7/9

To solve this question, you can solve for the probability of choosing 2 marbles that are pink and subtracting that from 1 to obtain the probability of selecting any variation of marbles that are not both pink.

The probability of picking 2 marbles that are both pink would be the product of the probability of choosing the first pink marble multiplied by the probability of choosing a second pink marble from the remaining marbles in the mix.

This would be 1/2 * 4/9 = 2/9.

To obtain the probability that is asked, simply compute 1 – (2/9) = 7/9.

The probability that the 2 randomly chosen marbles are not both pink is 7/9.

### Example Question #1 : Outcomes

Choose a number at random from 1 to 5.

Column A

The probability of choosing an even number

Column B

The probability of choosing an odd number

**Possible Answers:**

Column A is greater

Column A and B are equal

Cannot be determined

Column B is greater

**Correct answer:**

Column B is greater

There are two even numbers and three odd numbers, so P (even) = 2/5 and P (odd) = 3/5.

### Example Question #1 : How To Find The Probability Of An Outcome

Two fair dice are thrown. What is the probability that the outcome will either total 7 or include a 3?

**Possible Answers:**

2/3

8/9

7/12

5/12

1/2

**Correct answer:**

5/12

If a die is rolled twice, there are 6 * 6 = 36 possible outcomes.

Each number is equally probable in a fair die. Thus you only need to count the number of outcomes that fulfill the requirement of adding to 7 or including a 3. These include:

1 6

2 5

3 4

4 3

5 2

6 1

3 1

3 2

3 3

3 5

3 6

1 3

2 3

5 3

6 3

This is 15 possibilities. Thus the probability is 15/36 = 5/12.

### Example Question #1 : How To Find The Probability Of An Outcome

Box A has 10 green balls and 8 black balls.

Box B has 9 green balls and 5 black balls.

What is the probability if one ball is drawn from each box that both balls are green?

**Possible Answers:**

^{19}/_{252}

^{9}/_{14}

^{10}/_{49}

^{5}/_{14}

^{5}/_{9}

**Correct answer:**

^{5}/_{14}

Note that drawing balls from each box are independent events. Thus their probabilities can be combined with multiplication.

Probability of drawing green from A:

10/18 = 5/9

Probability of drawing green from B:

^{9}/_{14}

So:

^{5}/_{9} * ^{9}/_{14} = ^{5}/_{14}

### Example Question #1 : How To Find The Probability Of An Outcome

The probability that events A and/or B will occur is 0.88.

Quantity A: The probability that event A will occur.

Quantity B: 0.44.

**Possible Answers:**

Quantity B is greater.

The relationship cannot be determined from the information given.

The two quantities are equal.

Quantity A is greater.

**Correct answer:**

The relationship cannot be determined from the information given.

The only probabilites that we know from this is that P(only A) + P(only B) + P (A and B) = 0.88, and that P(neither) = 0.12. We cannot calculate the probability of P(A) unless we know two of the probabilites that add up to 0.88.

### Example Question #2 : How To Find The Probability Of An Outcome

a is chosen randomly from the following set:

{3, 11, 18, 22}

b is chosen randomly from the following set:

{ 4, 8, 16, 32, 64, 128}

What is the probability that a + b = 27?

**Possible Answers:**

0.04

0.03

0.5

0.1

0.05

**Correct answer:**

0.04

Since any of the first set can be summed with any of the second set, the addition sign in the equation works like a conjunction. As such, there are 4 * 6 = 24 possible combinations of a and b. Only one of these combinations, 11 + 16 = 27, works. Thus the probability is 1/24, or about 0.04.

### Example Question #8 : Probability

There are four aces in a standard deck of playing cards. What is the approximate probability of drawing two consecutive aces from a standard deck of 52 playing cards?

**Possible Answers:**

0.004

0.005

0.05

0.5

0.4

**Correct answer:**

0.005

Answer: .005

Explanation: The probability of two consecutive draws without replacement from a deck of cards is calculated as the number of possible successes over the number of possible outcomes, multiplied together for each case. Thus, for the first ace, there is a 4/52 probability and for the second there is a 3/51 probability. The probability of drawing both aces without replacement is thus 4/52*3/51, or approximately .005.

### Example Question #2 : How To Find The Probability Of An Outcome

In a bag, there are 10 red, 15 green, and 12 blue marbles. If you draw two marbles (without replacing), what is the approximate probability of drawing two different colors?

**Possible Answers:**

67.57%

0.06%

33.33%

25%

None of the other answers

**Correct answer:**

67.57%

Calculate the chance of drawing either 2 reds, two greens, or two blues. Then, subtract this from 1 (100%) to calculate the possibility of drawing a pair of different colors.

The combined probability of RR, GG, and BB is: (10 * 9) / (37 * 36) + (15 * 14) / (37 * 36) + (12 * 11) / (37 * 36)

This simplifies to: (90 + 210 + 132) / 1332 = 432 / 1332

Subtract from 1: 1 - 432 / 1332 = (1332 - 432) / 1332 = approx. 0.6757 or 67.57%

### Example Question #3 : How To Find The Probability Of An Outcome

What is the probability of drawing 2 hearts from a standard deck of cards without replacement?

**Possible Answers:**

13/52

12/52

1/16

1/17

1/4

**Correct answer:**

1/17

There are 52 cards in a standard deck, 13 of which are hearts

13/52 X 12/51 =

1/4 X 12/51 =

12/ 204 = 3/51 = 1/17