GRE_Math_Bible_eBook

Counting 411

7. Let A be the set of families having exactly one car. Then the question is how many families are there in set A.

Next, let B be the set of families having exactly two cars, and let C be the set of families having more than two cars.

Then the set of families having at least one car is the collection of the three sets A, B, and C.

The number of families in the three sets A, B, and C together is 250 (given) and the number of families in the two sets B and C together is 60 (given).

Now, since set A is the difference between a set containing the three families of A, B, and C and a set of families of B and C only, the number of families in set A equals

(the number of families in sets A, B, and C together) – (the number of families in sets B and C) = 250 – 60 =

190

The answer is (B).

8.Let the number of female children Emma has ben. Since Anna herself is one of them, she has n – 1 sisters. Hence, as given, she must have the same number (= n – 1) of brothers. Hence, the number of male children Emma has is n – 1. Since Andrew is one of them, Andrew has (n – 1) – 1 = n – 2 brothers. Now, the number of sisters Andrew has (includes Anna) is n (= the number of female children). Since Andrew has twice as many sisters as brothers, we have the equation n = 2(n – 2). Solving the equation for n yields n

=4. Hence, Emma has 4 female children, and the number of male children she has is n – 1 = 4 – 1 = 3. Hence, the total number of children Emma has is 4 + 3 = 7. The answer is (D).

9.The trainer wants to make teams of either 3 participants each or 5 participants each or 6 participants each successfully without leaving out any one of the participants in the batch. Hence, the batch size must be a multiple of all three numbers 3, 5, and 6. Hence, the batch size must be a multiple of the least common multiple of 3, 5, and 6, which is 30. The answer is (E).

10.There are three kinds of voters:

1)Voters who voted for A only. Let the count of such voters be a.

2)Voters who voted for B only. Let the count of such voters be b.

3)Voters who voted for both A and B. The count of such voters is 50 (given).

Since the total number of voters is 250, we have

a + b + 50 = 250

a + b = 200 (1) By subtracting 50 from both sides Now, we have that 100 voters voted for A. Hence, we have

(Voters who voted for A only) + (Voters who voted for both A and B) = 100 Forming this as an equation yields

a + 50 = 100 a = 50

Substituting this in equation (1) yields 50 + b = 200. Solving for b yields b = 150. The answer is (C).

414 GRE Math Bible

What’s the probability of drawing a blue marble and then a red marble from a bowl containing 4 red marbles, 5 blue marbles, and 5 green marbles? (Assume that the marbles are not replaced after being selected.) As calculated before, there is a 5/14 likelihood of selecting a blue marble first and a 4/13 likelihood of selecting a red marble second. Forming the product yields the probability of a red marble

immediately followed by a blue marble: 145 134 = 18220 = 1091.

These two examples can be generalized into the following rule for calculating consecutive probabilities:

To calculate consecutive probabilities, multiply the individual probabilities.

This rule applies to two, three, or any number of consecutive probabilities.

Either-Or Probabilities

What’s the probability of getting either heads or tails when flipping a coin once? Since the only possible outcomes are heads or tails, we expect the probability to be 100%, or 1: 12 + 12 = 1. Note that the events heads and tails are mutually exclusive. That is, if heads occurs, then tails cannot (and vice versa).

What’s the probability of drawing a red marble or a green marble from a bowl containing 4 red marbles, 5 blue marbles, and 5 green marbles? There are 4 red marbles out of 14 total marbles. So the probability of selecting a red marble is 4/14 = 2/7. Similarly, the probability of selecting a green marble is 5/14. So

exclusive. For instance, if a red marble is selected, then neither a blue marble nor a green marble is selected.

These two examples can be generalized into the following rule for calculating either-or probabilities:

To calculate either-or probabilities, add the individual probabilities (only if the events are mutually exclusive).

The probabilities in the two immediately preceding examples can be calculated more naturally by adding up the events that occur and then dividing by the total number of possible events. For the coin example, we get 2 events (heads or tails) divided by the total number of possible events, 2 (heads and tails): 2/2 = 1. For the marble example, we get 9 (= 4 + 5) ways the event can occur divided by 14 (= 4 + 5 + 5) possible events: 9/14.

If it’s more natural to calculate the either-or probabilities above by adding up the events that occur and then dividing by the total number of possible events, why did we introduce a second way of calculating the probabilities? Because in some cases, you may have to add the individual probabilities. For example, you may be given the individual probabilities of two mutually exclusive events and be asked for the probability that either could occur. You now know to merely add their individual probabilities.

Geometric Probability

In this type of problem, you will be given two figures, with one inside the other. You’ll then be asked what is the probability that a randomly selected point will be in the smaller figure. These problems are solved

with the same principle we have been using: Probability = desired outcome . possible outcomes

Example: In the figure to the right, the smaller square has sides of length 2 and the larger square has sides of length 4. If a point is chosen at random from the large square, what is the probability that it will be from the small square?

Probability & Statistics 415

STATISTICS

Statistics is the study of the patterns and relationships of numbers and data. There are four main concepts that may appear on the test:

Median

When a set of numbers is arranged in order of size, the median is the middle number. For example, the median of the set {8, 9, 10, 11, 12} is 10 because it is the middle number. In this case, the median is also the mean (average). But this is usually not the case. For example, the median of the set {8, 9, 10, 11, 17}

is 10 because it is the middle number, but the mean is 11 = 8 + 9 +10 +11+17 . If a set contains an even 5

number of elements, then the median is the average of the two middle elements. For example, the median

Example: What is the median of 0, –2, 256 , 18, 2 ?

Arranging the numbers from smallest to largest (we could also arrange the numbers from the largest to smallest; the answer would be the same), we get –2, 0, 2 , 18, 256. The median is the middle number,

2 .

Mode

The mode is the number or numbers that appear most frequently in a set. Note that this definition allows a set of numbers to have more than one mode.

Example: What is the mode of 3, –4, 3 , 7, 9, 7.5 ?

The number 3 is the mode because it is the only number that is listed more than once. Example: What is the mode of 2, , 2 , –9, , 5 ?

Both 2 and are modes because each occurs twice, which is the greatest number of occurrences for any number in the list.

Range

The range is the distance between the smallest and largest numbers in a set. To calculate the range, merely subtract the smallest number from the largest number.

Example: What is the range of 2, 8, 1 , –6, , 1/2 ?

The largest number in this set is 8, and the smallest number is –6. Hence, the range is 8 – (–6) = 8 + 6 = 14.

Standard Deviation

On the test, you are not expected to know the definition of standard deviation. However, you may be presented with the definition of standard deviation and then be asked a question based on the definition. To make sure we cover all possible bases, we’ll briefly discuss this concept.

Standard deviation measures how far the numbers in a set vary from the set’s mean. If the numbers are scattered far from the set’s mean, then the standard deviation is large. If the numbers are bunched up near the set’s mean, then the standard deviation is small.

Example: Which of the following sets has the larger standard deviation?

A = {1, 2, 3, 4, 5}

B = {1, 4, 15, 21, 27}

All the numbers in Set A are within 2 units of the mean, 3. All the numbers in SetB are greater than 5 units from the mean, 15. Hence, the standard deviation of Set B is greater.

416GRE Math Bible

Problem Set Z:

Easy

1.The minimum temperatures from Monday through Sunday in the first week of July in southern Iceland are observed to be –2°C, 4°C, 4°C, 5°C, 7°C, 9°C, 10°C. What is the range of the temperatures?

(A)–10°C

(B)–8°C

(C)8°C

(D)10°C

(E)12°C

Medium

2.What is the probability that the product of two integers (not necessarily different integers) randomly selected from the numbers 1 through 20, inclusive, is odd?

(A)0

(B)1/4

(C)1/2

(D)2/3

(E)3/4

3.Two data sets S and R are defined as follows:

Data set S: 28, 30, 25, 28, 27

Data set R: 22, 19, 15, 17, 21, 25

By how much is the median of data set S greater than the median of data set R?

(A)5

(B)6

(C)7

(D)8

(E)9

4.If x and y are two positive integers and x + y = 5, then what is the probability that x equals 1?

(A)1/2

(B)1/3

(C)1/4

(D)1/5

(E)1/6

5.The following values represent the number of cars owned by the 20 families on Pearl Street. 1, 1, 2, 3, 2, 5, 4, 3, 2, 4, 5, 2, 6, 2, 1, 2, 4, 2, 1, 1

What is the probability that a family randomly selected from Pearl Street has at least 3 cars?

(A)1/6

(B)2/5

(C)9/20

(D)13/20

(E)4/5

Probability & Statistics 417

6.The following frequency distribution shows the number of cars owned by the 20 families on Pearl Street.

12

22

3a

44

5 5

62

What is the probability that a family randomly selected from the street has at least 4 cars?

(A)1/10

(B)1/5

(C)3/10

(D)9/20

(E)11/20

7.Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirty-five envelopes in the bag are unstamped, and 5/7 of the unstamped envelopes are airmail letters. What is the probability that an envelope picked randomly from the bag is an unstamped airmail envelope?

(A)1/7

(B)1/3

(C)13/35

(D)17/38

(E)23/70

8.Set S is the set of all numbers from 1 through 100, inclusive. What is the probability that a number randomly selected from the set is divisible by 3?

(A)1/9

(B)33/100

(C)34/100

(D)1/3

(E)66/100

9.What is the probability that the sum of two different numbers randomly picked (without replacement) from the set S = {1, 2, 3, 4} is 5?

(A)1/5

(B)3/16

(C)1/4

(D)1/3

(E)1/2

10.The ratio of the number of red balls, to yellow balls, to green balls in a urn is 2 : 3 : 4. What is the probability that a ball chosen at random from the urn is a red ball?

(A)2/9

(B)3/9

(C)4/9

(D)5/9

(E)7/9

418GRE Math Bible

11. The frequency distribution for x is as given below. What is the range of f ?

x f

01

1 5

24

13. A prize of $200 is given to anyone who solves a hacker puzzle independently. The probability that Tom will win the prize is 0.6, and the probability that John will win the prize is 0.7. What is the probability that both will win the prize?

(A) 0.35

(B) 0.36

(C) 0.42

(D) 0.58

(E) 0.88

14. If the probability that Mike will miss at least one of the ten jobs assigned to him is 0.55, then what is the probability that he will do all ten jobs?

(A) 0.1

(B) 0.45

(C) 0.55

(D) 0.85

(E) 1

15. The probability that Tom will win the Booker prize is 0.5, and the probability that John will win the Booker prize is 0.4. There is only one Booker prize to win. What is the probability that at least one of them wins the prize?

(A) 0.2

(B) 0.4

(C) 0.7

(D) 0.8

(E) 0.9

420GRE Math Bible

20.Every one who passes the test will be awarded a degree. The probability that Tom passes the test is 0.3, and the probability that John passes the test is 0.4. The two events are independent of each other. What is the probability that at least one of them gets the degree?

(A)0.28

(B)0.32

(C)0.5

(D)0.58

(E)0.82

21.A national math examination has 4 statistics problems. The distribution of the number of students who answered the questions correctly is shown in the chart. If 400 students took the exam and each question was worth 25 points, then what is the average score of the students taking the exam?

1200

2304

3350

4250

(A)1 point

(B)25 points

(C)26 points

(D)69 points

(E)263.5 points

Very Hard

22.There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?

(A)6/14

(B)2/7

(C)6/35

(D)6/29

(E)6/42