GRE Math : GRE Quantitative Reasoning

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Example Questions

Example Question #481 : Gre Quantitative Reasoning

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.

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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% less likely

just as likely

80% less likely

20% more likely

80% more likely

Correct answer:

80% less likely

Explanation:

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.

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Example Question #25 : 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:

$\frac{4}{17}$

$\frac{3}{52}$

$\frac{5}{51}$

$\frac{1}{17}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{17}$

Explanation:

For the first card, the probability of choosing a diamond is $\dpi{100} \small \frac{13}{52}$ . With no replacing, the probability of choosing another diamond is $\dpi{100} \small \frac{12}{51}$ . Multilying these probabilities yields $\dpi{100} \small \frac{3}{51}$ .

This reduces to $\frac{1}{17}$ .

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Example Question #482 : Gre Quantitative Reasoning

You see someone correctly guess a card from a 52 card deck and you want to out-do him in an equally "odds-defying" feat. How many sequential coin flips would you have to predict to do something "less-likely" than guessing the correct card from a 52 card deck?

Possible Answers:

$\dpi{100} \small 4$

$\dpi{100} \small 10$

$\dpi{100} \small 5$

$\dpi{100} \small 8$

$\dpi{100} \small 6$

Correct answer:

$\dpi{100} \small 6$

Explanation:

You are trying to have "worse" odds than a $\dpi{100} \small \frac{1}{52}$ chance. 5 correct coin flip guesses would be a $\dpi{100} \small (\frac{1}{2})^{5} = \frac{1}{32}$ chance, not quite as unlikely.

6 correct guesses however would be a $\dpi{100} \small (\frac{1}{2})^{6} = \frac{1}{64}$ chance, thus "outdoing" the odds of the card guess.

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Example Question #483 : Gre Quantitative Reasoning

A fair six-sided die is thrown.

Quantity A: The probability that the die will show either a "6" or an odd number.

Quantity B: $\dpi{100} \small \frac{2}{3}$

Possible Answers:

Quantity B is greater.

The two quantities are equal.

Quantity A is greater.

The relationship cannot be determined from the information given.

Correct answer:

The two quantities are equal.

Explanation:

To obtain probability, you must divide the number of possible desired outcomes by the number of total possible outcomes. The number of desired outcomes is 4 (3 odd numbers plus 1 even number). The number of total outcomes is 6 (number of sides on the die). So $\dpi{100} \small \frac{4}{6}$ reduces to $\dpi{100} \small \frac{2}{3}$ , making the 2 quantities equal.

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Example Question #484 : Gre Quantitative Reasoning

What is the probability of rolling a total of 5 with a single roll of two fair 6-sided dice, each with the distinct numbers 1 through 6 on each side?

Possible Answers:

$\dpi{100} \small \frac{1}{6}$

$\dpi{100} \small \frac{1}{7}$

$\dpi{100} \small \frac{1}{9}$

$\dpi{100} \small \frac{2}{7}$

$\dpi{100} \small \frac{1}{3}$

Correct answer:

$\dpi{100} \small \frac{1}{9}$

Explanation:

$\dpi{100} \small Probability=\frac{\&hash;\ of\ desired\ outcomes}{\&hash;\ of\ possible\ outcomes}$

When one die is rolled, there are six possible outcomes, when two dice are rolled, there are $\dpi{100} \small 6\times 6=36$ possible outcomes. Getting a total value of 5 can be achieved in the following ways:

$\dpi{100} \small (1,4), (2,3), (3,2),(4,1)$

So four possible ways. So the probability of rolling a total of 5 is $\dpi{100} \small \frac{4}{36}$ , which is $\dpi{100} \small \frac{1}{9}$ .

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Example Question #484 : Gre Quantitative Reasoning

A fair six-sided die is thrown.

Quantity A: The probability that the die will show either an even number or a 1

Quantity B: $\frac{2}{3}$

Possible Answers:

The information cannot be determined from the information given

Quantity A is greater

The two quantities are equal

Quantity B is greater

Correct answer:

The two quantities are equal

Explanation:

A die has six sides and there are 3 possible even numbers that the die could land on. Thus the probability that the die will land on an even number (2,4,6) or a 1 is $\frac{4}{6}$ , or $\frac{2}{3}$ .

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Example Question #485 : Gre Quantitative Reasoning

A six-sided die is thrown twice. What is the probability that both numbers are 5 or 6?

Possible Answers:

$\small \frac{1}{6}$

$\small \frac{1}{18}$

$\small \frac{1}{3}$

$\small \frac{1}{2}$

$\small \frac{1}{9 }$

Correct answer:

$\small \frac{1}{9 }$

Explanation:

The probability of each die rolling a 5 or 6 is $\small \frac{1}{3}$ . The probability of both rolls being 5 or 6 is $\small \frac{1}{3}*\frac{1}{3}=\frac{1}{9}$ .

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Example Question #36 : Probability

In a regular deck of cards, what is the probability that a card drawn at random will be a diamond or a face card?

Aces are not a face card.

Possible Answers:

$\small \frac{25}{104}$

$\small \frac{25}{52}$

$\small \frac{11}{52}$

$\small \frac{3}{52}$

$\small \frac{11}{26}$

Correct answer:

$\small \frac{11}{26}$

Explanation:

In a regular deck, there are 13 cards of each suit, so 13 diamonds. Also, there are four cards of each rank, so the set of face cards contains 12 cards in total (4 kings, 4 queens, and 4 jacks).

If these sets of diamonds and face cards did not overlap, there would be 13 + 12 = 25 cards of interest to choose; however, three face cards are also diamonds (king of diamonds, queen of diamonds, jack of diamonds), so we need to subtract 3 from the combined set to avoid double-counting, giving a total of 22 cards of interest.

Finally, to get the probability, we divide the number of cards of interest by the total number of cards in the deck.

$\frac{22}{52}=\frac{11}{26}$

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Example Question #486 : Gre Quantitative Reasoning

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?

Possible Answers:

$\frac{3}{7}$

$\frac{3}{15}$

$\frac{7}{15}$

Correct answer:

Explanation:

The probability is 1 because since there are 15 people and only 7 days of the week.

$\frac{15}{7}=2\ R1$

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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Example Question #37 : Probability

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?

Possible Answers:

Correct answer:

Explanation:

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: $p(A\ or\ B)=0.22+0.35=0.57$

Not mutually exclusive: $p(A\ or\ B)=0.22+0.35-p(A\ and\ B)$

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).

$p(A\ or\ B)=0.54=0.22+0.35-p(A\ and\ B)=0.57-p(A\ and\ B)$

$p(A\ and\ B)=0.03$

Now we can solve to ratio of p(A and B) to p(C).

p(C)=0.33

$\frac{p(C))}{p(A\ and\ B)}=\frac{0.33}{0.03}=11$

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GRE Math : GRE Quantitative Reasoning

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #481 : Gre Quantitative Reasoning

Example Question #25 : Probability

Example Question #482 : Gre Quantitative Reasoning

Example Question #483 : Gre Quantitative Reasoning

Example Question #484 : Gre Quantitative Reasoning

Example Question #484 : Gre Quantitative Reasoning

Example Question #485 : Gre Quantitative Reasoning

Example Question #36 : Probability

Example Question #486 : Gre Quantitative Reasoning

Example Question #37 : Probability

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