All GRE Math Resources
Example Questions
Example Question #1 : How To Find The Probability Of An Outcome
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.
The relationship cannot be determined from the information given.
The quantities are equal.
Quantity A is greater.
Quantity B is greater.
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 #4 : Data Analysis
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?
2/9
7/9
7/8
5/6
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
Column A is greater
Cannot be determined
Column B is greater
Column A and B are equal
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?
5/12
8/9
2/3
7/12
1/2
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 : 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?
5/9
9/14
10/49
5/14
19/252
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.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A is greater.
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?
0.5
0.03
0.04
0.05
0.1
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 #2 : 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?
0.4
0.005
0.05
0.5
0.004
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 #4 : 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?
0.06%
33.33%
None of the other answers
25%
67.57%
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 #1 : Probability
What is the probability of drawing 2 hearts from a standard deck of cards without replacement?
1/17
13/52
1/4
12/52
1/16
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