GMAT Math : Problem-Solving Questions

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Example Questions

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 #1541 : Psat Mathematics

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

Possible Answers:

$\frac{1}{128}$

$\frac{6}{7}$

$\frac{127}{128}$

$\frac{1}{7}$

$\frac{255}{256}$

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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Example Question #164 : Data Analysis

In a particular high school, 200 students are freshmen, 150 students are sophomores, 250 students are juniors, and 100 students are seniors. Twenty percent of freshmen are in honors classes, ten percent of sophomores are in honors classes, twelve percent of juniors are in honors classes, and thirty percent of seniors are in honors classes.

If a student is chosen at random, what is the probability that that student will be a student who attends honors classes?

Possible Answers:

$\frac{5}{7}$

$\frac{9}{50}$

$\frac{9}{100}$

$\frac{23}{70}$

$\frac{23}{140}$

Correct answer:

$\frac{23}{140}$

Explanation:

First calculate the number of students:

200+150+250+100=700

The probability of drawing an honors student will then be the total number of honors students divided by the total number of students attending the school:

$\frac{200(.20)+(150)(.10)+(250)(.12)+(100)(0.30)}{700}$

$\frac{115}{700}=\frac{23}{140}$

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Example Question #61 : Discrete Probability

If a student is chosen at random, what is the probability that that student will be a senior student and a student who does not attend honors classes?

Possible Answers:

$\frac{3}{10}$

$\frac{4}{35}$

$\frac{2}{5}$

$\frac{3}{70}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{10}$

Explanation:

First calculate the number of students:

200+150+250+100=700

The percentage of seniors that do not attend honors classes is:

(1-.30)=.70

Therefore, the probability of selecting a student who is a senior and one who does not attend honors classes is:

$\frac{100}{700}(.70)=\frac{1}{10}$

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Example Question #61 : Data Analysis

A card is drawn at random from a deck of fifty-two cards (no jokers). What is the probability of drawing a diamond, a card with an even number, or a card with a number divisible by three?

Possible Answers:

$\frac{45}{52}$

$\frac{33}{52}$

$\frac{17}{26}$

$\frac{1}{2}$

$\frac{19}{26}$

Correct answer:

$\frac{17}{26}$

Explanation:

First, consider the probability of each individual event: drawing a diamond, an even number, or a number divisible by three.

Drawing a diamond, there are four suits, so:

$P(D)=\frac{1}{4}$

Drawing an even number, there are five possible values [2,4,6,8,10] present in each thirteen-card suit, so:

$P(E)=\frac{5}{13}$

Drawing a number divisible by three, there are three possible values [3,6,9] present in each thirteen-card suit, so

$P(3)=\frac{3}{13}$

This problem deals with a union of probabilities, essentially a "this or that" option; however, since there are three events considered, the formula for the union follows the form:

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

The probability of drawing an even number that is also a diamond is:

$\frac{5}{52}$

The probability of drawing a diamond that is divisible by three is:

$\frac{3}{52}$

The probability of drawing an even number that is divisible by three is:

$\frac{1}{13}$

And the probability of drawing a number that is a diamond, even, and divisble by three is:

$\frac{1}{52}$

Therefore, the probability of this union is:

$\frac{13+20+12-5-3-4+1}{52}=\frac{34}{52}=\frac{17}{26}$

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Example Question #1851 : Gmat Quantitative Reasoning

Some balls are placed in a large box, which include one ball marked "A", two balls marked "B", three balls marked "C", and so forth up to twenty-six balls marked "Z".

A ball is drawn at random. What is the probability that the ball will be marked with a letter in the word "roost"?

Possible Answers:

$\frac{72}{325}$

$\frac{8}{78}$

$\frac{89}{234}$

$\frac{89}{117}$

$\frac{8}{39}$

Correct answer:

$\frac{8}{39}$

Explanation:

The total number of balls in the box will be

1 + 2 + 3 + ... + 26 .

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 26 = \frac{26 (26 + 1)}{2} = \frac{26 \cdot 27}{2} = 351$ .

There are four distinct letters in the word "roost" - "O", "R", "S", and "T", the fifteenth, eighteenth, nineteenth, and twentieth letters of the alphabet. Therefore, there will be a total of 15+18+19 + 20 = 72 balls with one of these letters. The probability of drawing one of them is

$\frac{72}{351} = \frac{72 \div 9 }{351\div 9 } = \frac{8}{39}$

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GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #44 : Probability

Example Question #41 : Probability

Example Question #1541 : Psat Mathematics

Example Question #1548 : Psat Mathematics

Example Question #52 : Calculating Discrete Probability

Example Question #42 : Data Analysis / Probablility

Example Question #164 : Data Analysis

Example Question #61 : Discrete Probability

Example Question #61 : Data Analysis

Example Question #1851 : Gmat Quantitative Reasoning

All GMAT Math Resources

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