GRE Math : Probability

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Example Questions

Example Question #41 : How To Find The Probability Of An Outcome

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?

Possible Answers:

$0.15$

$0.21$

$0.11$

$0.13$

$0.2$

Correct answer:

$0.11$

Explanation:

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 $\frac{7}{16}$ , the second draw of $\frac{6}{15}$ , and the third draw of $\frac{9}{14}$ .

$\frac{7}{16}\times \frac{2}{5}\times \frac{9}{14}=\frac{9}{80}\approx 0.11$

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Example Question #49 : Data Analysis

There is a bag of 28 red balls and a bag of 12 blue balls.

Quantity A: The total number of red and blue balls.

Quantity B: The number of blue balls that you can pair with one red ball.

Possible Answers:

Quantity A is greater

The relationship cannot be determined from the information given

Quantity B is greater

The two quantities are equal

Correct answer:

Quantity A is greater

Explanation:

There are 40 total balls, and only 12 blue balls.

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Example Question #491 : Gre Quantitative Reasoning

A random variable is normally distributed with a mean of 500 and a standard deviation of 20.

---

Probability of the event that x<450

---

$\frac{1}{10}$

Possible Answers:

Quantity A is greater

The two quantities are equal

Quantity B is greater

The relationship cannot be determined

Correct answer:

Quantity B is greater

Explanation:

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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Example Question #51 : Calculating Discrete Probability

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

Possible Answers:

$\frac{25}{52}$

$\frac{9}{26}$

$\frac{11}{26}$

$\frac{3}{52}$

$\frac{156}{2704}$

Correct answer:

$\frac{11}{26}$

Explanation:

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 $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

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 $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

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Example Question #44 : Probability

A coin is flipped four times. What is the probability of getting heads at least three times?

Possible Answers:

$\frac{1}{4}$

$\frac{1}{16}$

$\frac{11}{16}$

$\frac{5}{16}$

$\frac{3}{4}$

Correct answer:

$\frac{5}{16}$

Explanation:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

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Example Question #41 : Probability

Rolling a four-sided dice, what is the probability of rolling a three times out of four?

Possible Answers:

$\frac{3}{64}$

$\frac{13}{256}$

$\frac{3}{256}$

$\frac{243}{256}$

$\frac{3}{128}$

Correct answer:

$\frac{3}{64}$

Explanation:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case rolling a four), and is the probability of success (one in four).

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{4})^3(1-\frac{1}{4})^{4-3}=4(\frac{1}{64})(\frac{3}{4})=\frac{3}{64}$

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Example Question #42 : How To Find The Probability Of An Outcome

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

Possible Answers:

$\frac{1}{128}$

$\frac{1}{7}$

$\frac{255}{256}$

$\frac{127}{128}$

$\frac{6}{7}$

Correct answer:

$\frac{127}{128}$

Explanation:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

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Example Question #1548 : Psat Mathematics

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is picked from Set A and Set B. What is the probability of picking two consonants?

Possible Answers:

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{3}{14}$

$\frac{5}{14}$

$\frac{2}{14}$

Correct answer:

$\frac{5}{14}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:

$\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

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Example Question #52 : Calculating Discrete Probability

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is picked from Set A and Set B. What is the probability of picking at least one consonant?

Possible Answers:

$\frac{5}{13}$

$\frac{1}{7}$

$\frac{5}{14}$

$\frac{6}{7}$

$\frac{17}{14}$

Correct answer:

$\frac{6}{7}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:

$P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)$

The interesection is:

$P(A\bigcap B)=P(A)P(B)=\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

So, we can find the probability of drawing at least one consonant:

$P(A\bigcup B)= \frac{5}{7}+\frac{1}{2}-\frac{5}{14}=\frac{10}{14}+\frac{7}{14}-\frac{5}{14}=\frac{6}{7}$

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Example Question #42 : Data Analysis / Probablility

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is drawn from Set A, and one from Set B. What is the probability of drawing a matching pair of letters?

Possible Answers:

$\frac{1}{42}$

$\frac{4}{13}$

$\frac{1}{21}$

$\frac{2}{13}$

$\frac{1}{7}$

Correct answer:

$\frac{1}{21}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

Between Set A and Set B, there are two potential matching pairs of letters: AA and XX. The amount of possible combinations is the number of values in Set A, multiplied by the number of values in Set B, 7(6)=42 .

Therefore, the probability of drawing a matching set is:

$\frac{2}{42}=\frac{1}{21}$

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GRE Math : Probability

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #41 : How To Find The Probability Of An Outcome

Example Question #49 : Data Analysis

Example Question #491 : Gre Quantitative Reasoning

Example Question #51 : Calculating Discrete Probability

Example Question #44 : Probability

Example Question #41 : Probability

Example Question #42 : How To Find The Probability Of An Outcome

Example Question #1548 : Psat Mathematics

Example Question #52 : Calculating Discrete Probability

Example Question #42 : Data Analysis / Probablility

All GRE Math Resources

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