GMAT Math : Discrete Probability

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Example Question #71 : Calculating Discrete Probability

Target

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

A blindfolded woman throws a dart at random at the above target. Let be the probability that the dart lands inside a green region; let and be similarly defined for a red or yellow region. Which of the following is a true statement?

(Assume that skill is not a factor, and that the dart hits the target.)

Possible Answers:

Y < R < G

Y < R =G

G = Y< R

Y< G < R

G < Y< R

Correct answer:

Y < R < G

Explanation:

Since the areas of the sectors are what are important here, we need only rank the total areas of the different colors, which we will call G', R', Y' .

For the sake of simplicity, we will assume the smaller semicircle of the target has radius ; consequently, the larger semicircle has radius .

The larger 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 smaller semicircle has total area

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

Each of the six sectors of the larger semicircle has area $\frac{1}{6}$ of its area, or

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

Each of the two sectors of the smaller semicircle has area $\frac{1}{2}$ of its area, or

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

The total of the areas of the green sectors - three sectors of the larger semicircle - is

$G' = 3 \cdot \frac{1}{3} \pi = \pi$

The total of the areas of the red sectors - two sectors of the larger semicircle and one of the smaller - is

$R' = 2 \cdot \frac{1}{3} \pi + \frac{1}{4} \pi = \frac{2}{3} \pi + \frac{1}{4} \pi = \frac{8}{12} \pi + \frac{3}{12} \pi = \frac{11}{12} \pi$

The total of the areas of the yellow sectors - one sector of the larger semicircle and one of the smaller - is

$Y' = \frac{1}{3} \pi + \frac{1}{4} \pi = \frac{4}{12} \pi + \frac{3}{12} \pi = \frac{7}{12} \pi$

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

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Example Question #71 : Discrete Probability

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. Give the odds against the ball being marked "I".

Possible Answers:

$\textup{78 to 1}$

$\textup{39 to 1}$

$\textup{76 to 1}$

$\textup{38 to 1}$

$\textup{77 to 1}$

Correct answer:

$\textup{38 to 1}$

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 "I" is the ninth letter of the alphabet, there are nine balls marked "I", and 351-9 = 342 other balls. Therefore, the odds against drawing a ball marked "I" are 342 to 9, or

$\frac{342}{9} = \frac{342 \div 9}{9 \div 9} = \frac{38}{1}$

The correct choice is 38 to 1.

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Example Question #73 : Calculating Discrete Probability

Target

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

A blindfolded woman throws a dart at random at the above target. Disregarding any skill factor, and assuming that the dart hits the target, what are the odds against the dart landing inside the purple region?

Possible Answers:

$\textup{11 to 1}$

$\textup{13 to 2}$

$\textup{7 to 1}$

$\textup{6 to 1}$

$\textup{11 to 2}$

Correct answer:

$\textup{13 to 2}$

Explanation:

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

The larger 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 smaller 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

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

The purple sector has area one-sixth that of the larger semicircle, or

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

The area that is not purple is

$\frac{5}{2} \pi - \frac{1}{3} \pi = \frac{15}{6} \pi - \frac{2}{6} \pi = \frac{13}{6} \pi$

The odds against the dart landing in the purple region are

$\frac{\frac{13}{6} \pi}{\frac{1}{3} \pi} = \frac{\frac{13}{6} \pi \cdot \frac{6}{\pi}}{\frac{1}{3} \pi \cdot \frac{6}{\pi}} = \frac{13}{2}$ - that is, 13 to 2.

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Example Question #72 : Discrete Probability

Eduardo has a standard deck of cards plus two jokers (54 cards in all). What are the odds that he draws a joker, followed by a red card that isn't a king? Assume replacement and that there is one red joker and one black joker.

Possible Answers:

0.17

0.017

0.184

0.083

Correct answer:

0.017

Explanation:

This is a probability question. Probability can be found via the following:

$P=\frac{\#\ of\ desired\ outcomes}{total\ number\ of\ outcomes}$

To find the probability of multiple events, we find the probability of each event and multiply them together.

The probability of the first event is as follows:

$P_1=\frac{2}{54}=\frac{1}{27}$

Note that there are 54 cards and only two jokers.

The second event will still have the same denominator as the first event. The numerator however, will need to change.

"...a red card that isn't a king..." So, our deck would normally be half red. However, two of those cards will be kings, so the number of desired outcomes should be 25

$P_2=\frac{25}{54}$

Next, we need to multiple our two probabilities together.

$P_{tot}=\frac{25}{54}*\frac{1}{27}=\frac{25}{1458}\approx0.017$

So our answer is:

0.017

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Example Question #75 : Calculating Discrete Probability

Some balls are placed in a large box; the balls include one ball marked "A", two balls marked "B", and so forth up to twenty-six balls marked "Z". A ball is drawn at random.

Given a particular letter of the alphabet, does the probability that that ball will be marked with that letter exceed $\frac{1}{30}$ ?

Statement 1: The letter appears in the word "lousy".

Statement 2: The letter appears in the word "skunk".

Possible Answers:

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

Correct answer:

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

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

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

The number of balls with a given letter of the alphabet is equal to the number of its position in the alphabet; the probability of a ball with that letter being drawn is that number divided by the total number of balls, 351. Therefore, for this probability to exceed $\frac{1}{30}$ , we must have the relation

$\frac{M}{351} > \frac{1}{30}$

$\frac{M}{351} \cdot 351 > \frac{1}{30} \cdot 351$

$M > \frac{351}{30} = 11 \frac{7}{10}$ .

Therefore, M > 11 .

The 11th letter of the alphabet is "K", so in order to answer this question, it suffices to know whether the letter comes after "K" in the alphabet.

From Statement 1 alone, the question can be answered, since all of the letters in the word "lousy" appear after "K" in the alphabet. From Statement 2 alone, however, the question cannot be answered, since the letter "K" itself appears in the word "skunk".

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Example Question #311 : Arithmetic

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

If the above spinner is spun, what are the odds against the arrow stopping in the purple region?

Possible Answers:

13 to 2

7 to 1

6 to 1

11 to 2

11 to 1

Correct answer:

11 to 1

Explanation:

The radii of the portions 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 purple sector is one-sixth of the larger semicircle, so it is one-twelfth of a circle. This means that the probability of a spinner stopping inside that sector is $\frac{1}{12}$ , and the odds against this are

$\frac{\frac{11}{12} }{\frac{1}{12}} = \frac{11}{1}$ - that is, 11 to 1.

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Example Question #77 : Calculating Discrete Probability

Some balls are placed in a large box; the balls include one ball marked "A", two balls marked "B", and so forth up to twenty-six balls marked "Z". A ball is drawn at random.

Given a particular letter of the alphabet, does the probability that that ball will be marked with that letter exceed $\frac{1}{20}$ ?

Statement 1: The letter in question is in the word "pique".

Statement 2: The letter in question is a vowel.

(Note: for purposes of this problem, "Y" is a consonant.)

Possible Answers:

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

Correct answer:

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

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

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

$\frac{M}{351} > \frac{1}{20}$

$\frac{M}{351} \cdot 351 > \frac{1}{20} \cdot 351$

$M > \frac{351}{20} = 17 \frac{11}{20}$ .

Therefore, M > 17 .

The 17th letter of the alphabet is "Q", so in order to answer this question, it suffices to know whether the letter comes after "Q" in the alphabet.

Assume both statements to be true. There are three vowels in "pique"; "E" and "I" come before "Q" and "U", after "Q". Therefore, the two statements together are inconclusive.

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Example Question #81 : Probability

Aaron, Gary, Craig, and Boone are sitting down in a row of four chairs. What is the probability that Aaron and Gary will be seated beside each other?

Possible Answers:

$\frac{11}{12}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{3}{4}$

$\frac{13}{24}$

Correct answer:

$\frac{1}{2}$

Explanation:

Consider first all of the possible ways the men may be arranged, which is

4!=24

Now, consider all of the ways that Aaron and Gary could be seated beside each other; it may be easier to visualize by drawing it out:

A G _ _
G A _ _
_ A G _
_ G A _
_ _ A G
_ _ G A

As seen, there are six possibilities.

Finally, for each of these cases, Craige and Boone could be seated in one of two ways.

So the probability that Aaron and Gary will be seated beside each other is:

$\frac{2\cdot 6}{24}=\frac{12}{24}=\frac{1}{2}$

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GMAT Math : Discrete Probability

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #71 : Calculating Discrete Probability

Example Question #71 : Discrete Probability

Example Question #73 : Calculating Discrete Probability

Example Question #72 : Discrete Probability

Example Question #75 : Calculating Discrete Probability

Example Question #311 : Arithmetic

Example Question #77 : Calculating Discrete Probability

Example Question #81 : Probability

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