GMAT Math : Calculating discrete probability

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Example Questions

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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Example Question #303 : Arithmetic

Target

The upper half of the above target has radius twice that of the lower half.

A blindfolded man throws a dart at random at the above target. Disregarding any skill factor, and assuming that the dart hits the target, what is the probability that it will land inside a red region?

Possible Answers:

$\frac{5}{12}$

$\frac{3}{8}$

None of the other responses gives the correct answer.

$\frac{7}{18}$

$\frac{11}{30}$

Correct answer:

$\frac{11}{30}$

Explanation:

For the sake of simplicity, we will assume the smaller half of the target has radius 1, and, consequently, the larger half has radius 2.

The upper semicircle has total area

$A_{1} = \frac{1}{2} \cdot \pi r_{1}^{2} = \frac{1}{2} \cdot \pi \cdot 2 ^{2} = \frac{1}{2} \cdot 4 \pi = 2 \pi$

The lower semicircle has total area

$A_{2} = \frac{1}{2} \cdot \pi r_{2} ^{2} = \frac{1}{2} \cdot \pi \cdot 1 ^{2} = \frac{1}{2} \pi$

The total area of the target is

$A = A_{1}+ A_{2} = 2 \pi + \frac{1}{2} \pi = \frac{5}{2} \pi$

Each of the six sectors of the upper semicircle has area one sixth of its area, or

$\frac{ 1}{6}\cdot 2 \pi = \frac{1}{3} \pi$

Each of the two sectors of the lower semicircle has area one half of its area, or

$\frac{ 1}{2}\cdot \frac{ 1}{2} \pi = \frac{1}{4} \pi$

The red regions comprise two sectors of the upper semicircle and one of the lower, so their total area is

$\frac{1}{3} \pi + \frac{1}{3} \pi + \frac{1}{4} \pi = \frac{4}{12} \pi + \frac{4}{12} \pi + \frac{3}{12} \pi = \frac{11}{12} \pi$

The probability of the dart hitting one of these red regions is therefore

$\frac{ \frac{11}{12} \pi}{\frac{5}{2} \pi} = \frac{ \frac{11}{12} \pi}{\frac{30}{12} \pi} = \frac{11}{30}$

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Example Question #304 : Arithmetic

The upper semicircle of the above spinner has radius twice that of the lower semicircle.

If the above spinner is spun, what is the probability that the arrow will stop inside a yellow region?

Possible Answers:

$\frac{1}{5}$

$\frac{5}{18}$

$\frac{1}{3}$

$\frac{7}{30}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{3}$

Explanation:

The radii of the semicircles of the spinner - and the areas of the sectors - are actually irrelevant to the problem; it is the measures of their central angles that count.

The yellow regions comprise one sector of the top semicircle and one sector of the bottom one. The former is one-sixth of a semicircle and measures $\frac{1}{6} \times 180^{\circ } = 30^{\circ }$ ; the latter is a fourth of a circle and measures $\frac{1}{4} \times 360^{\circ } = 90^{\circ }$ . Therefore, the total of the measures of the central angles is

$30^{\circ }+ 90 ^{\circ }= 150^{\circ }$ ,

and the probability that the arrow will stop in one of these regions is

$\frac{120}{360} = \frac{120 \div 120 }{360\div 120 } = \frac{1}{3}$ .

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Example Question #311 : Arithmetic

Some balls are placed in a large box, which include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1".

What is the probability that a ball randomly drawn from this box will be marked with an even number?

Possible Answers:

$\frac{5}{9}$

$\frac{5}{11}$

$\frac{6}{11}$

$\frac{1}{2}$

$\frac{4}{9}$

Correct answer:

$\frac{5}{11}$

Explanation:

The total number of balls in the box will be

1 + 2 + 3 + ... + 10 .

Since

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

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The even numbers are 2, 4, 6, 8, and 10. According to the pattern, there are nine balls marked "2", seven marked "4", five marked "6", three marked "8", and one marked "10". Therefore, there are

9+5+7+3+1 = 25

balls marked with even numbers.

The probability of drawing a ball with an even number is therefore

$\frac{25}{55} = \frac{5}{11}$

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Example Question #312 : Arithmetic

Some balls are placed in a large box, which include one ball marked , two balls marked , and continuing up to balls marked .

What is the probability that a ball randomly drawn from this box will be marked with a prime number?

Possible Answers:

$\frac{2}{5}$

$\frac{18}{55}$

$\frac{1}{2}$

$\frac{17}{55}$

$\frac{17}{45}$

Correct answer:

$\frac{17}{55}$

Explanation:

The total number of balls in the box will be

1 + 2 + 3 + ... + 10 .

Since,

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

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The prime numbers in the range of $\textup{1 to 10}$ are $\textup{ 2, 3, 5, and 7}$ . The number of balls with each number corresponds to the number itself, so the number of balls with prime numbers will be

2+3+5+7 = 17 .

The probability of drawing one of them is $\frac{17}{55}$ .

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Example Question #313 : Arithmetic

Spinner 2

The upper semicircle of the above spinner has radius twice that of the lower semicircle.

Suppose the arrow is spun. Let be the probability that the arrow stops when pointing to a green region; let and be similarly defined for a red or yellow region. Which of the following is a true statement?

Possible Answers:

Y < R = G

Y < R < G

R = G < Y

Y < G< R

G < Y < R

Correct answer:

G < Y < R

Explanation:

The radii of the halves of the spinner - and the areas of the sectors - are actually irrelevant to the problem; it is the measures of their central angles that count. In order to compare the probabilities of the arrow stopping in regions of the three colors, it suffices to compare the totals of the measures of the central angles of the different colors - we will call them G', R', Y' .

Each sector of the larger semicircle is one-sixth of a $180^{\circ }$ sector and has central angle

$\frac{1}{6} \times 180^{\circ } = 30^{\circ }$ .

Each sector of the smaller semicircle is a quarter of a circle, and has central angle

$\frac{1}{4} \times 360^{\circ } = 90^{\circ }$ .

The green regions comprise three sectors of the top semicircle, so the total of their central angles is

$G' = 3 \times 30^{\circ } = 90^{\circ }$ .

The yellow regions comprise one sector of the top semicircle and one sector of the bottom one, so the total of their central angles is

$Y' =30^{\circ } +90^{\circ } = 120^{\circ }$ .

The red regions comprise two sectors of the top semicircle and one sector of the bottom one, so the total of their central angles is

$R' = 2 \times 30^{\circ } + 90 ^{\circ } = 60^{\circ } +90^{\circ } = 150^{\circ }$ .

G'<Y'<R' ; it follows that G<Y<R .

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Example Question #1861 : Problem Solving Questions

Some balls are placed in a large box, which include ball marked "A", balls marked "B", balls marked "C", and so on up to balls marked "Z".

A ball is drawn at random. Give the probability that the ball will be marked "O".

Possible Answers:

$\frac{5}{117}$

$\frac{3}{130}$

$\frac{3}{65}$

$\frac{3}{70}$

$\frac{5}{234}$

Correct answer:

$\frac{5}{117}$

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

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

Since "O" is the fifteenth letter of the alphabet, there are balls out of 351 marked with "O". Therefore, the probability that a randomly-drawn ball will be one of these is

$\frac{15}{351} = \frac{15\div 3 }{351 \div 3 } = \frac{5}{117}$

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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 #164 : Data Analysis

Example Question #61 : Discrete Probability

Example Question #61 : Data Analysis

Example Question #1851 : Gmat Quantitative Reasoning

Example Question #303 : Arithmetic

Example Question #304 : Arithmetic

Example Question #311 : Arithmetic

Example Question #312 : Arithmetic

Example Question #313 : Arithmetic

Example Question #1861 : Problem Solving Questions

All GMAT Math Resources

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