# GRE Math : Data Analysis

## Example Questions

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

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

Column A is greater

Column A and B are equal

Column B is greater

Cannot be determined

Column B is greater

Explanation:

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?

7/12

5/12

8/9

2/3

1/2

5/12

Explanation:

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 : Probability

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?

10/49

5/9

9/14

5/14

19/252

5/14

Explanation:

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 : Probability

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.

The two quantities are equal.

The relationship cannot be determined from the information given.

Quantity B is greater.

Quantity A is greater.

The relationship cannot be determined from the information given.

Explanation:

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 #11 : Data Analysis

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?

0.05

0.03

0.1

0.5

0.04

0.04

Explanation:

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 #1 : How To Find The Probability Of An Outcome

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?

0.05

0.4

0.5

0.005

0.004

0.005

Explanation:

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 : Outcomes

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?

33.33%

25%

67.57%

0.06%

67.57%

Explanation:

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 #1 : How To Find The Probability Of An Outcome

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

1/16

1/17

12/52

13/52

1/4

1/17

Explanation:

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

### Example Question #11 : Outcomes

Mike has a bag of marbles. 5 are green, 8 are red, and 3 are blue. He pulls one marble out of the bag and it is green. He pulls out another one and it is red. He does not return these marbles to the bag. What is the probability that the next marble he pulls out of the bag will be green?

5/14

4/16

3/16

4/14

4/14

Explanation:

First we need to find out how many marbles are in the bag in total. 5 + 8 + 3 = 16. He removes a green marble so now there are only 15 in total. When he removes the red one there are then 14 marbles in the bag. 14 is your denominator. The odds of picking a green one are 5 – 1 or 4 because there are only 4 green marbles left in the bag; therefore, the odds of picking another green marble is 4/14

### Example Question #12 : How To Find The Probability Of An Outcome

A classroom has 9 boys and 9 girls. One student is chosen at random as the class leader, and a second student is chosen at random as a back-up leader.

Quantity A: The probability of choosing a boy to be the leader and choosing a girl to be the back-up

Quantity B: The probability of choosing boys for both roles

Quantity A is greater

The relationship cannot be determined from the information give

The quantities are equal

Quantity B is greater

Quantity A is greater

Explanation:

There are 18 students in the class, and 2 must be selected from the 9 girls and 9 boys. Key to this question is noting that the 2 students must be unique: ie once a student is selected to lead the class, he or she cannot be chosen to be the back-up.

Since these are independent events, the probability of each event is found, and the events are multiplied by each other to find the total.

Quantity A:

P(boy leader) = 9/18 = 1/2 as there are 9 boys out of a possible 18 students.

Once the boy has been chosen, there are 8 boys and 17 students total from which to choose the second student.

P(girl back-up) = 9/17 because there are 9 girls and 17 students left.

P(Quantity A) = (1/2)(9/17) = 9/34

Quantity B:

P(boy leader) = 9/18 = 1/2 as there are 9 boys out of a possible 18 students.

Once the boy has been chosen, there are 8 boys and 17 students total from which to choose the second student.

P(boy back-up) = 8/17 because there are 8 boys and 17 students left.

P(Quantity B) = (1/2)(8/17) = 8 / 34 = 4/17

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