### All GRE Math Resources

## Example Questions

### Example Question #21 : Probability

A high school has 200 students. 120 are male, 50 are upper division students, and 40 are upper division male students. What is the probability of choosing a lower division female student, given the student is female?

**Possible Answers:**

2/5

7/8

7/20

7/15

1/4

**Correct answer:**

7/8

There are 200 students in total, and 120 of them are male, so 80 must be female. We also know that there are 50 upper division students, and 40 of them are male, so 10 must be female. If 10 of 80 females are upper division, the other 70 females have to be lower division students, so the probability of choosing a lower division student, given the student is female, is 70/80 = 7/8.

### Example Question #21 : How To Find The Probability Of An Outcome

Out of 25 dresses, 9 have defects. Of those 9 defective dresses, 6 have holes and 3 have rips. What is the probability that a randomly selected defective dress has a rip, given that it has a defect?

**Possible Answers:**

3/5

1/3

1/9

4/9

2/3

**Correct answer:**

1/3

We are given that the chosen dress has a defect, so we are looking at the 9 defective dresses. 3 of the defective dresses have rips, so the probability that a dress has a rip, given that is has a defect, is 3/9 = 1/3.

### Example Question #22 : How To Find The Probability Of An Outcome

If you throw a die three times, what is the probability of getting a 1 on at least one of the three throws?

**Possible Answers:**

1/3

125/216

25/36

91/216

5/6

**Correct answer:**

91/216

Let's first find the probability of not getting a 1 on any of the three throws. Prob(no 1 on first throw) = 5/6. We need Prob(no 1 on first throw AND no 1 on second throw AND no 1 on third throw). "And" signifies multiplication in probability, so Prob(no 1 on all three throws) = 5/6 * 5/6 * 5/6 = 125/216. Now, to find the probability of getting a 1 on at least one throw, we simply take 1 – Prob(no 1 on all three throws) = 1 – 125/216 = 91/216.

### Example Question #21 : Probability

Every day is either rainy or sunny. Mondays are rainy with probability 3/5. Tuesdays are sunny with probability 1/4. Wednesdays are also sunny with probability 2/3. What is the probability that the weather is the same on Monday, Tuesday, and Wednesday?

**Possible Answers:**

15/19

1/8

3/20

1/15

13/60

**Correct answer:**

13/60

We want either rainy, rainy, rainy or sunny, sunny, sunny.

P(rain on Mon AND rain on Tues AND rain on Wed) = 3/5 * 3/4 * 1/3 = 9/60

P(sun on Mon AND sun on Tues AND sun on Wed) = 2/5 * 1/4 * 2/3 = 4/60

P(same weather) = P(all 3 rainy days OR all 3 sunny days)

= P(all 3 rainy days) + P(all 3 sunny days)

= 9/60 + 4/60 = 13/60.

### Example Question #33 : Data Analysis

16 dogs and 21 cats are up for adoption at the pound. 12 of these animals are brown and 19 are female. What is the probability of randomly selecting a brown animal?

**Possible Answers:**

12/19

16/19

12/37

19/37

Cannot be determined

**Correct answer:**

12/37

The problem gives us extra information. We know 12 animals are brown, and there are 16 + 21 = 37 animals in total. Then the probability of choosing a brown animal is simply 12/37.

### Example Question #21 : How To Find The Probability Of An Outcome

What is the probability of pulling 2 spades out of a standard deck of cards without replacement?

**Possible Answers:**

**Correct answer:**

There are 52 cards in a standard deck. 13 of them are spades.

So there is a chance of pulling out the first spade. That leaves 51 cards and 12 spades left. Now there is a chance to get a second spade.

When you combine probabilities, multiply the two individual probabilites together.

### Example Question #21 : Probability

What is the probability of fliiping a coin 4 times and getting 2 heads.

**Possible Answers:**

**Correct answer:**

There are 4 flips of the coin. chance that the first is a heads, the second is a heads, the third is a tails, and the fourth is a tails. There are 6 permutations that give a combined 2 heads and 2 tails in whatever order.

### Example Question #21 : Probability

A bag contains 4 blue marbles, 2 black marbles, 1 clear marble, and 3 red marbles. What is the probability of pulling out 3 red marbles without replacement?

**Possible Answers:**

**Correct answer:**

There are 10 marbles in the bag. So, the first probability of pulling out the first red marble is ; the next is ; the final is .

Multiply the individual probabilities together to find the total outcome.

### Example Question #21 : Probability

A study was conducted to determine the effectiveness of a vaccine for the common cold (Rhinovirus sp.). 1000 patients were studied. Of those, 500 received the vaccine and 500 did not. The patients were then exposed to the Rhinovirus and the results were tabulated.

Table 1 shows the number of vaccinated and unvaccinated patients in each age group who caught the cold.

In order to determine the effectiveness of the vaccine, scientists want to show the impact of the vaccine on a patient's odds of catching a cold.

A patient who receives the vaccination is __________ to catch Rhinovirus than an unvaccinated patient.

**Possible Answers:**

20% more likely

just as likely

80% more likely

80% less likely

20% less likely

**Correct answer:**

80% less likely

According to the table, 50 unvaccinated patients caught Rhinovirus.

18 + 4 + 5 + 4 + 19 = 50

250 unvaccinated patients caught Rhinovirus.

63 + 32 + 29 + 51 + 75 = 250

Next, we need to determine the percentage relationship between the number of unvaccinated versus vaccinated patients that caught the virus.

50/250 = 20%

100% – 20% = 80%

Because only 20% as many vaccinated patients exposed to the cold virus caught the cold, this reflects an 80% reduction in the likelihood of a patient to catch the cold after receiving the vaccine.

### Example Question #22 : Probability

A standard deck of cards contains 52 cards and 13 of each suit. Megan chooses two cards, without replacing. What is the probability that Megan chooses two diamonds?

**Possible Answers:**

**Correct answer:**

For the first card, the probability of choosing a diamond is . With no replacing, the probability of choosing another diamond is . Multilying these probabilities yields .

This reduces to .