Outcomes - GRE Quantitative Reasoning
Card 1 of 608
What is the probability of drawing 2 hearts from a standard deck of cards without replacement?
What is the probability of drawing 2 hearts from a standard deck of cards without replacement?
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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
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
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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?
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?
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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:
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
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
The chance of BOTH of these selections being made is then the product of the two probalities:
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What is the probability of drawing 2 hearts from a standard deck of cards without replacement?
What is the probability of drawing 2 hearts from a standard deck of cards without replacement?
Tap to reveal answer
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
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
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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?
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?
Tap to reveal 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:
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
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
The chance of BOTH of these selections being made is then the product of the two probalities:
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There are 15 people in a room. What is the probablility that at least 3 of them were born on the same day of the week?
There are 15 people in a room. What is the probablility that at least 3 of them were born on the same day of the week?
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The probability is 1 because since there are 15 people and only 7 days of the week.
This means even if each of the 7 days only had 2 people's birthday on them after checking the first 14 people the 15th and final person must also have a birthday on one of these days, so at the very least one day of the week will include at least 3 people's birthday.
The probability is 1 because since there are 15 people and only 7 days of the week.
This means even if each of the 7 days only had 2 people's birthday on them after checking the first 14 people the 15th and final person must also have a birthday on one of these days, so at the very least one day of the week will include at least 3 people's birthday.
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The probability that event A occurs is 0.22. The probability of event B occuring is 0.35. If the probability of event C occuring is 0.33, and the probability of event A or event B occuring is 0.54, then the probability of event C occuring is how many times the probability of event A and B occuring?
The probability that event A occurs is 0.22. The probability of event B occuring is 0.35. If the probability of event C occuring is 0.33, and the probability of event A or event B occuring is 0.54, then the probability of event C occuring is how many times the probability of event A and B occuring?
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Keywords in this question are or and and. Here, the probability of event A or B is p(A) + p(B) if they are mutually exclusive. If they are not, however, the probability of A or B is p(A) + p(B) - p(A and B).
Mutually exclusive: 
Not mutually exclusive: 
Given that p(A) + p(B) = 0.57, and we are told that p(A or B) = 0.54, it implies that p(A and B) = 0.03 (and that A and B are not mutually exclusive).
Now we can solve to ratio of p(A and B) to p(C).
Keywords in this question are or and and. Here, the probability of event A or B is p(A) + p(B) if they are mutually exclusive. If they are not, however, the probability of A or B is p(A) + p(B) - p(A and B).
Mutually exclusive:
Not mutually exclusive:
Given that p(A) + p(B) = 0.57, and we are told that p(A or B) = 0.54, it implies that p(A and B) = 0.03 (and that A and B are not mutually exclusive).
Now we can solve to ratio of p(A and B) to p(C).
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Two six-sided dice are rolled.
Quantity A: The probability of rolling a total of 6
Quantity B: The probability of rolling a total of 9
Two six-sided dice are rolled.
Quantity A: The probability of rolling a total of 6
Quantity B: The probability of rolling a total of 9
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List out the the possibilities of combinations for rolling a total of 6 and the possibilities of combinations for rolling a total of 9. There are 5 possible outcomes for rolling a total of 6 and 4 possible outcomes for rolling a total of 9. There are a total of 36 possible outcomes, so the probability of rolling a 6 is 
and the possibility of rolling a 9 is 
.
List out the the possibilities of combinations for rolling a total of 6 and the possibilities of combinations for rolling a total of 9. There are 5 possible outcomes for rolling a total of 6 and 4 possible outcomes for rolling a total of 9. There are a total of 36 possible outcomes, so the probability of rolling a 6 is
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Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?
Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?
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A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is 
.
There are thirteen spades in the deck, so the probability of drawing a spade is 
.
Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is 
Thus the probability of drawing a face card or a spade is:
A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is
There are thirteen spades in the deck, so the probability of drawing a spade is
Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is
Thus the probability of drawing a face card or a spade is:
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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.
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.
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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.
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.
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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?
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?
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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.
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.
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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
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
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There are two even numbers and three odd numbers, so P (even) = 2/5 and P (odd) = 3/5.
There are two even numbers and three odd numbers, so P (even) = 2/5 and P (odd) = 3/5.
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Two fair dice are thrown. What is the probability that the outcome will either total 7 or include a 3?
Two fair dice are thrown. What is the probability that the outcome will either total 7 or include a 3?
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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.
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.
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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?
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?
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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
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
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There are seven black marbles and nine white marbles in a bag. What is the approximate probability of drawing two black marbles and then a white marble without replacement?
There are seven black marbles and nine white marbles in a bag. What is the approximate probability of drawing two black marbles and then a white marble without replacement?
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This is a straightforward probability problem. Recall that to find the probability of related draws without replacement, we multiply the relative probabilities of each event. The first draw has a probability of 
, the second draw of 
, and the third draw of 
.
This is a straightforward probability problem. Recall that to find the probability of related draws without replacement, we multiply the relative probabilities of each event. The first draw has a probability of
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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 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.
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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.
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.
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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?
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?
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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.
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.
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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?
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?
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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.
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.
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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?
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?
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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%
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%
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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?
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?
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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
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
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A random variable 
is normally distributed with a mean of 500 and a standard deviation of 20.
A
---
Probability of the event that 
B
---
A random variable
A
---
Probability of the event that
B
---
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In a normally distributed curve, a standard deviation of 1 contains ~68% of all values within its range, and a standard deviation of 2 contains ~95% of all values within its range. Since the mean value is 500 for this situation, two standard deviations would occur at values 460 to 540 (i.e. ~95% of all values are within this range). This would mean that the value of 450 would fall outside this range, i.e. a 100%–95% = 5% probability. Since 5% expressed as a fraction is 1/20, Quantity B is far greater than the probability of landing at 450.
In a normally distributed curve, a standard deviation of 1 contains ~68% of all values within its range, and a standard deviation of 2 contains ~95% of all values within its range. Since the mean value is 500 for this situation, two standard deviations would occur at values 460 to 540 (i.e. ~95% of all values are within this range). This would mean that the value of 450 would fall outside this range, i.e. a 100%–95% = 5% probability. Since 5% expressed as a fraction is 1/20, Quantity B is far greater than the probability of landing at 450.
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