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.
Quantity B is greater.
The quantities are equal.
The relationship cannot be determined from the information given.
Quantity A 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 #1 : Probability
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?
7/8
7/9
5/6
2/9
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 : Probability
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 A and B are equal
Column B is greater
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?
8/9
1/2
5/12
2/3
7/12
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 #2 : 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?
9/14
5/14
10/49
5/9
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.
The relationship cannot be determined from the information given.
Quantity A is greater.
The two quantities are equal.
Quantity B 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?
0.05
0.03
0.04
0.1
0.5
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 #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
0.004
0.005
0.4
0.5
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 #1 : Probability
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?
None of the other answers
33.33%
0.06%
67.57%
25%
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?
13/52
1/16
1/4
1/17
12/52
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