GMAT-Arithmetic-4. Counting Methods and Probability

Transcription

1 GMAT-Arithmetic-4 Counting Methods and Probability Counting Methods: 1).A new flag with six vertical stripes is to be designed using some or all of the colours yellow, green, blue and red. The number of ways this can be done such that no two adjacent stripes have same colour is : A B C D E. Cannot be determined Since no adjacent stripes contain the same color. The second, and the remaining vertical stripe will have only three options, while the first will have any of the four options. 4*3*3*3*3*3 = 12* 3^4 = 12 * 81. So the answer is A. 2).There are 6 tasks and 6 persons. Task1 cannot be assigned either to person 1 or to person 2. Task2 must be assigned either to person 3 or to person 4. Every person is to be assigned one task. In how many ways can this assignment be done? A. 144 B. 180 C. 192 D. 360 E. 716 First fill the positive condition then the negative condition. 1

2 Task 2 must be assigned to person 3 or to person 4: So Task 2 has only 2 options. Task1 cannot be assigned either to person 1 or to person 2: So Task1 has only 3 options (Because task 2 must have been assigned to anyone of the person 3 or person 4). So there are remaining 4 tasks and 4 persons so can be assigned 4! Ways. Task1 Task2 Task3 Task4 Task5 Task6 3 * 2 * 4 * 3 * 2 * 1 = 6 * 4! =6 *24. = 144. So the answer is A. 3). There are 20 players in a certain chess tournament. If each of the players is to play each of the other players exactly once, how many total matches will be played? A. 20 B. 21 C. 40 D. 190 E. 20! To pick the first team the number of ways is: 20 To pick the second team the number of ways is: 19 However the arrangement of the teams doesn t matter hence the total number of ways: (20 x 19)/2 = 190 The question can also be solved in the following manner. A chess match can occur between any 2 of the 20 players. So we just have to select 2 out of 20 and make them play a match. The number of ways of doing that is 20C2 =190. 2

3 So the answer is D. 4). A secretary differentiates files by a variety of color combinations. If she uses 4 different colors, and marks each file with 1, 2, or 3 of the colors, how many different color combinations can she generate? A. 4 B. 12 C. 14 D. 15 E Color out of Colors + 3 Colors 4C1 + 4C2 + 4C3 = 4 + (4 x 3)/2 + (4) = = 14. So the answer is C. 5).In how many ways can 3 Canadians,2 Americans and an Indian be arranged in a row so tha t the 3 Canadians are always together and the two Americans are always together? A. 360 B. 216 C. 72 D. 60 E. 36 Consider the 3 Canadians and 2 Americans as a single entity as they always are together. So we have the following pattern CAI (Canadian,American,Indian) which can be arranged in (3!) ways = 6 ways. Further the 3 Americans and 2 Canadians can be arranged among themselves in 3! and 2! Ways (among/between them). So required no of arrangements = 6 * 6 * 2 = 72 ways So the answer is C 6). If 5 points are indicated on a line and 4 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points? 3

4 A. 20 B. 30 C. 40 D. 70 E. 90 Let us say that there are two lines A and B. Line A has 5 points and line B has 4 points. To draw a triangle we need three points. We can choose three points in the following ways to make a triangle: A(2 points)b(1 Point) + A (1 Point) B(2 Points) 5c2 x 4c1 + 4c2 x 5c1 = = 70 7). How many 4-digit even numbers are there with distinct digits? A B C D E This question is about distinct digits. There are two cases, either zero can come at the end or 2, 4, 6 and 8 will come at the end. Both the cases are giving different possibilities. Let us explore both the cases: _ 0 + _ (2, 4, 6 or 8) Case 1: 9 x 8 x 7 x 1 We have to use zero at the fourth place and there are 9 numbers to choose for three places without any restrictions. 4

5 Case 2 : 8 x8 x 7 x 4 Total = 9 x 8 x 7 x x 8 x 7 x 4 = So the answer is B. 8). There are 5 students and 3 teachers. In how many ways can a team of 5 be formed so that there is at least one teacher but not more than two teachers in the team? A. 450 B. 180 C. 60 D. 45 E. 30 Team of 5 has to be formed with atleast one teacher and not more than two teachers. So there are two possibilities. One teacher 4 students (OR) two teacher 3 students. = 3C1 * 5C4 + 3C2 * 5C3. = 3 * * 10 = = 45. 9).How many diagonals are there in an octagon? A. 56 B. 28 C. 20 D. 16 E. 8 Diagonal is joining any two vertices other than sides. When we join any two points we get diagonal or sides. So there are 8 points in a octagon. 5

6 Joining any points means 8C2 = 28. This 28 equal to. 28 = number of Diagonals + number of Sides. We know the number of sides in a octagon is eight. So there are 20 Diagonals can be formed in Octagon. So the answer is C. 10).How many parallelograms are formed when a set of five parallel lines intersect anothe r set of four parallel lines? A. 120 B. 60 C. 36 D. 20 E. 10 Here we need to choose 2 parallel lines along with 2 vertical lines to form parallelogram. 2 parallel lines can be chosen from 5 parallel lines in 5C2 ways. 2 vertical(parallel) lines can be chosen from 4 vertical lines in 4C2 ways Together can be chosen to form a parallelogram in 5C2 * 4C2 ways = 10 * 6= 60. So the answer is B. Probability: 1).On average, a sharpshooter hits the target once every 3 shots. What is the probability that he will hit the target within 4 shots? A. 1 B. 1/81 C. 1/3 D. 65/81 6

7 E. 80/81 P(Hitting) = 1/3 P (Not Hitting) = 1 1/3 = 2/3 Probability of not hitting the target in four shots = (2/3) (2/3) (2/3) (2/3) = 16/81 P(Hitting in four shots) = 1 P(Not hitting in four shots) P(Hitting in four shots) = 1 (16/81) = 65/81. So the answer is D. 2).A bag contains 3 blue, 4 maroon and 2 orange balls. What is the probability of drawing a blue and an orange ball in two successive draws, each ball being put back after it is drawn? A. 2/27 B. 1/9 C. 1/3 D. 4/27 E. 2/9 3 Blue, 4 Maroon and 2 Orange = Total of 9 balls Probability of Complex Events = Product of Probability of individual events (Arrangement) Blue and Orange = (3/9) (2/9) (2!) = 4/27. So the answer is D. 3). How many times was a fair coin tossed? (1) If the coin had been tossed 4 times fewer, the probability of getting heads on every toss would have been 1/8. (2) When a coin is tossed this number of times, the number of different possible sequences of heads and tails is

8 Let us say that the coin was tossed n times. Statement I is sufficient Tossed = (n-4) times 1/8 correspond to HHH i.e. the coin was tossed 3 times N 4 = 3 N = 7 Statement II is sufficient 128 is only possible when there are seven coins (H or T) = 2 possible combinations one coin 2^7 combinations for 7 coins Hence answer is D. 4).If an integer p is to be chosen at random from 1 to 96 (both inclusive), what is the probability that p(p+1)(p+2) will be divisible by 8? A. ¼ B. 3/8 C. ½ D. 5/8 E. ¾ P = 1 Integer = 1 x 2 x 3 = 6 P = 2 Integer = 2 x 3 x 4 = 2 P = Even Integer = Even (Odd) (Even) = Divisible by 8 8

9 P = Odd Integer = Odd (Even) (Odd) = Only Divisible when this even number is divisible by 8 All the integers which are divisible by 8 between 1 and 96 = {8,16,24,32,40,48,56,64,72,80,88,96} P = Odd Values = {7,15,23,31,39,47,55,63,71,80,87,95} Favorable cases: All Even numbers + 12 Odd numbers = = 60 Probability = 60/96 = 5/8. So the answer is D. 5).Company A has 400 employees and Company B has 500 employees. Among these employees, there are 50 married couples, each consisting of an employee from A and an employee from B. If 1 employee is to be selected at random from each company, what is the probability that the 2 employees selected will be a married couple? A. 1 / 80 B. 7 / 9600 C. 1 / 4000 D. 1 / 192 E. 7 / 4800 Let us say that we choose a married employee from Company A: 50 ways to do that From Company B we have to choose his or her husband or wife: 1 way to do that Total number of ways: 50 x 1 Probability = 50 x 1 /(400 x 500) = 1/4000. So the answer is C. 6).Three numbers from 1 to 100 (inclusive) are selected at random and the numbers can be repeated. What is the probability that the sum of the three numbers will be odd? A. 1/4 9

10 B. 3/8 C. 1/2 D. 5/8 E. ¾ For sum of three numbers to be odd the three numbers can be: {Odd, Odd, Odd} OR {Even, Even, Odd} From 1 to 100 Probability of picking up an Odd Number is 1/2 which is same as the probability of picking up Even numbers: Odd, Odd, Odd: (1/2) (1/2) (1/2) = 1/8 Even, Even, Odd: (1/2) (1/2) (1/2) (3) = 3/8 (We multiplied by three as Even, Even and Odd can arrange in three different ways) Probability = 1/8 + 3/8 = 4/8 = ½. So the answer is C. 7).In a group of 8 persons what is the probability that at least two of them are born on the same day of the week? A. 1/8 B. 1/7 C. 1/4 D. 1/2 E. 1 Let A,B,C,D,E,F,G and H be the members of the group. Let us distribute evenly as shown below Sunday - A Monday - B Tuesday - C 10

11 Wednesday - D Thursday - E Friday - F Saturday - G and H should be born on any of these days. So the probability that at least TWO of the group members are born on the same day of the week is 1. 8).In a drawer of shirts, 8 are blue, 6 are green and 4 are magenta. If Mason draws 2 shirts at random, what is the probability that at least one of the shirts he draws will be blue? A. 25/153 B. 23/153 C. 5/17 D. 4/9 E. 12/17 There are a total of 18 shirts. 8 blue and 10 non blue. P(selecting at least 1 blue shirt) = 1 - P(selecting no blue shirts) P(selecting first non-blue shirt) = 10/18 P(selecting secong non-blue shirt) = 9/17 P(selecting no blue shirts) = 10/18 * 9/17= 10/34 Therefore P(selecting at least 1 blue shirt) = 1 -(10/34)= 24/34=12/17. So the answer is E. 9).Probability of A attending class is 1/6, probability of B attending class is 1/5. What is the probability that either A alone or B alone attends class? A. 1/10 B. 1/5 C. 3/10 D. 1/3 11

12 E. 11/30 Probability of A attending the class is 1/6. Probability of A not attending the class is 5/6. Probability of B attending the class is 1/5. Probability of B not attending the class is 4/5. Either A alone or B alone attending the class = P(A attending)*p(b not attending)+ P(B attending)*p(a not attending) = 1/6 * 4/5 + 1/5 * 5/6 = 9/30 = 3/10. So the answer is C. 10).Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws? A. 1/(5^4) B. 1/(5^3) C. 6/(5^4) D. 13/(5^4) E. 17/(5^4) Given P(S) = 1/5, so P(F) = 4/5 At least 3 out of 4 means 3 or 4. That is she has having success in 3 throws or 4 throws. 12

13 Success in 3 throws means = SSSF =P(SSSF) = (1/5) *(1/5)* (1/5)* (4/5) * 4!/3!(arrangement)= (4/5^4) * 4 = 16/5^4. Success in 4 throws means= SSSS =P(SSSS) = (1/5) *(1/5)* (1/5)* (1/5) * 4!/4!(arrangement)= (1/5^4) * 1 = 1/5^4. So adding both.(3 throws + 4 throws) =16/5^4 + 1/5^4 = 17/5^4. So the answer is E. 13