GMAT Math : Calculating discrete probability

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Example Questions

Example Question #1822 : Problem Solving Questions

In a 10,000 students university, in students will drop out of college. What is the probability that a randomly selected student will not drop out of college?

Possible Answers:

$\frac{3}{4}$

$\frac{3}{5}$

$\frac{1}{4}$

$\frac{2}{5}$

Correct answer:

$\frac{3}{4}$

Explanation:

The probability of students who will drop out is 1/4.

Therefore, we are looking for the probability that a student will NOT drop out, which is 3/4.

A randomly selected student will not drop out of college with a probability of 3/4.

$p=1-\frac{1}{4}=\frac{3}{4}$

The fact that there is 10,000 students in the school is not relevant to calculate this probability.

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Example Question #1823 : Problem Solving Questions

Of 100 students accepted for a masters degree program this fall semester, 30 do not have any work experience, 10 have two years of work experience, 15 have four years of work experience, and all of the other students have over seven years of work experience. What is the probability that a randomly selected student in the 100 who have been accepted for the program will have at least four years of work experience?

Possible Answers:

0.15

0.6

0.45

0.4

0.3

Correct answer:

0.6

Explanation:

The number of students who have over seven years of experience is:

100-30-10-15=100-55=45

The probability that a randomly selected student has at least four years of experience is:

Formula

Substituting in our values for these variables, we get:

$\frac{15+45}{100}=\frac{60}{100}=0.6$

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Example Question #1821 : Problem Solving Questions

An auto insurer underwrites its 60 customers and classifies them in 3 mutually exclusive risk classes. 15 of the customers are in the high-risk class, 35 are in the moderate-risk class, and 10 are in the low-risk class. What is the probability that a randomly selected customer will be in the moderate- or high-risk class?

Possible Answers:

$\frac{2}{3}$

$\frac{1}{6}$

$\frac{7}{48}$

$\frac{1}{2}$

$\frac{5}{6}$

Correct answer:

$\frac{5}{6}$

Explanation:

The probability that a randomly selected customer is in the moderate- or high-risk class is simply the sum of the number of clients in the moderate-risk class and the number of clients in the high-risk class divided by the total number of clients:

$\frac{15+35}{60}=\frac{50}{60}=\frac{5}{6}$

This is also equivalent to just summing up the probability of being in the high-risk class and the probability of being in the moderate-risk class. Since the 3 classes are mutually exclusive, we do not have to worry about subtracting the probability of mutual elements.

$\frac{15}{60}+\frac{35}{60}=\frac{5}{6}$

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Example Question #31 : Calculating Discrete Probability

A university received 1000 applications for their MBA program this semester. Of all applicants, 150 had a GMAT score above 720. 620 applicants had a GMAT score between 640 and 720. 80 applicants had a score between 560 and 640. All other applicants had a score lower than 560. What is the probability that a randomly selected applicant will have a score lower than 560?

Possible Answers:

0.15

0.10

0.12

0.08

0.16

Correct answer:

0.15

Explanation:

The probability that a randomly selected applicant has a score lower than 560 is:

$\frac{1000-150-620-80}{1000}=\frac{150}{1000}=0.15$

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Example Question #31 : Discrete Probability

Scientists tried out a new way of testing people for a certain disease. Out of 200 patients tested, tested positive when they did not have the disease. On the other hand, tested negative when they in fact had the disease. What is the probability that a randomly selected patient was accurately tested (i.e positive if he or she actually had the disease and negative if he or she did not)?

Possible Answers:

$\frac{17}{20}$

$\frac{1}{4}$

$\frac{5}{17}$

$\frac{3}{17}$

$\frac{11}{20}$

Correct answer:

$\frac{17}{20}$

Explanation:

Out of the 200 patients, wrongly tested positive and wrongly tested negative. There is a total of patients whose tests were not accurate.

The probability that a randomly selected patient was accurately tested is:

$\frac{200-30}{200}=\frac{170}{200}=\frac{17}{20}$

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Example Question #1831 : Problem Solving Questions

John alters a fair coin so that the probability of it coming up heads is 5% greater than that of it coming up tails. By how much has he increased the probability of it coming up heads?

Possible Answers:

$\frac{1}{41}$

$\frac{1}{82}$

$\frac{1}{8 1}$

$\frac{1}{40}$

$\frac{1}{80}$

Correct answer:

$\frac{1}{82}$

Explanation:

The probablilty of a fair coin coming up heads is $\frac{1}{2}$ .

If the coin is altered as in the problem, then the odds of the coin coming up heads as opposed to tails are 105 to 100; restated, the probability of it coming up heads is

$\frac{105}{205} = \frac{21}{41}$

John has increased the probability of the coin coming up heads by

$\frac{21}{41} - \frac{1}{2} = \frac{42}{82} - \frac{41}{82} = \frac{1}{82}$

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Example Question #32 : Calculating Discrete Probability

Marianna inserts the joker into a standard deck of 52 cards. By how much has she decreased the probability that a card drawn at random from the deck will be an ace?

Possible Answers:

$\frac{1}{689}$

$\frac{105}{2,756}$

$\frac{1}{53}$

$\frac{1}{2,756}$

$\frac{1}{52}$

Correct answer:

$\frac{1}{689}$

Explanation:

The probability that a card, randomly drawn from a standard deck of 52 cards without the joker, is an ace (or any other given rank) is $\frac{4}{52}= \frac{1}{13}$ ; if the joker is added, the probability decreases to $\frac{4}{53}$ . The decrease in probability is

$\frac{1}{13} - \frac{4}{53}=\frac{53 \cdot 1}{53 \cdot 13} - \frac{4\cdot 13}{53 \cdot 13} =\frac{53 }{689} - \frac{52}{689} = \frac{1}{689}$

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Example Question #33 : Calculating Discrete Probability

Pilar inserts the joker into a standard deck of 52 cards. By how much has she decreased the probability that a card drawn at random from the deck will be a diamond?

Possible Answers:

$\frac{1}{212}$

$\frac{105}{2,756}$

None of the other responses gives the correct answer.

$\frac{1}{53}$

$\frac{1}{2,756}$

Correct answer:

$\frac{1}{212}$

Explanation:

The probability that a card, randomly drawn from a standard deck of 52 cards without the joker, is a diamond (or any other given suit) is $\frac{13}{52}= \frac{1}{4}$ ; if the joker is added, the probability decreases to $\frac{13}{53}$ . The decrease in probability is

$\frac{1}{4} - \frac{13}{53}=\frac{53 \cdot 1}{53 \cdot 4} - \frac{13\cdot 4}{53 \cdot 4} =\frac{53 }{212} - \frac{52}{212} = \frac{1}{212}$ .

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Example Question #34 : Calculating Discrete Probability

Judy alters a fair six-sided die so that a roll of 1, 2, 3, 4, or 5 are still equally likely, but a roll of 6 is 5% more likely than any one of the other rolls.

By how much has she increased the probability of rolling a 6?

Possible Answers:

$\frac{1}{120}$

$\frac{5}{726}$

$\frac{1}{121}$

$\frac{7}{720}$

$\frac{1}{144}$

Correct answer:

$\frac{5}{726}$

Explanation:

The probability of rolling a 6 with a fair die is $\frac{1}{6}$ .

Since a roll of 6 is 5% more likely than any one of the other rolls, and the other rolls are equally probable, then, theoretically, there should be 105 rolls of 6 for every 100 rolls of 5, 100 rolls of 4, etc. - that is, 105 out of 605, rolls should theoretically be a 6. This makes the probability of rolling a 6 on the altered die

$\frac{105}{605} = \frac{21}{121}$ .

Therefore, Judy's alteration has increased the probability of rolling a 6 by

$\frac{21}{121} - \frac{1}{6} = \frac{21\cdot 6}{121 \cdot 6} - \frac{121 \cdot 1}{121 \cdot 6}= \frac{126}{726} - \frac{121 }{726} = \frac{5}{726}$

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Example Question #1836 : Gmat Quantitative Reasoning

Carla alters a fair six-sided die so that a roll of 1, 2, 3, 4, or 5 are still equally likely, but a roll of 6 is 4% more likely than any one of the other five outcomes.

By how much has she decreased the probability of rolling a 5?

Possible Answers:

$\frac{1}{906}$

$\frac{5}{906}$

$\frac{1}{150}$

$\frac{1}{151}$

$\frac{1}{604}$

Correct answer:

$\frac{1}{906}$

Explanation:

The probability of rolling a 5 with a fair die is $\frac{1}{6}$ .

Since a roll of 6 on the altered die is 4% more likely than any one of the other rolls, and the other rolls are equally probable, then, theoretically, there should be 104 rolls of 6 for every 100 rolls of 5, 100 rolls of 4, etc. Consequently, 100 out of 604 rolls should theoretically result in a 5. This makes the probability of rolling a 5 on the altered die

$\frac{100}{604} = \frac{25}{151}$ .

Therefore, Pilar's alteration has decreased the probability of rolling a 5 by

$\frac{1 }{6 } - \frac{25}{151} = \frac{1\cdot 151 }{6\cdot 151 } - \frac{6\cdot 25}{6\cdot 151} =\frac{ 151 }{906 } - \frac{150}{906} =\frac{1}{906}$

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GMAT Math : Calculating discrete probability

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1822 : Problem Solving Questions

Example Question #1823 : Problem Solving Questions

Example Question #1821 : Problem Solving Questions

Example Question #31 : Calculating Discrete Probability

Example Question #31 : Discrete Probability

Example Question #1831 : Problem Solving Questions

Example Question #32 : Calculating Discrete Probability

Example Question #33 : Calculating Discrete Probability

Example Question #34 : Calculating Discrete Probability

Example Question #1836 : Gmat Quantitative Reasoning

All GMAT Math Resources

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