Test: Probability- 2


25 Questions MCQ Test Quantitative Aptitude for GMAT | Test: Probability- 2


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QUESTION: 1

A pack of 12 cards contains either red or black cards. What is the probability that a card chosen at random will be a red card?

(1) The pack has 6 more red cards than black cards

(2) The pack has 3 red cards for every black card

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Given: A pack of 12 cards – Black or Red  

  • Let number of red cards = R
  • So, number of Black cards = 12 – R

To find: P(Choosing a Red card)

  • = R12
  • So, in order to answer the question, we need to find the value of R

Step 3: Analyze Statement 1 independently

Statement 1 says that ‘The pack has 6 more red cards than black cards

  • This means, R = (Number of Black cards)  + 6
  • R = (12 – R) + 6
  • 2R = 18
  • R = 9

Thus, Statement 1 alone is sufficient to provide a unique value of R

 

Step 4: Analyze Statement 2 independently

Statement 2 says that ‘The pack has 3 red cards for every black card’

  • NumberofRedCardsNumberofBlackcards=31
  •  
  • R12−R=31
  •  
  • R = 3(12 – R)
  • R = 36 – 3R
  • 4R = 36
  • R = 9

Thus, Statement 2 alone is sufficient to provide a unique value of R

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already got a unique answer in each of Steps 3 and 4, this step is not required

Answer: Option D

QUESTION: 2

Micky and Kevin go to a market to buy fruits. The probability of Micky buying an apple is 0.25 and of Kevin buying an apple is 0.4. Both Micky and Kevin have a 60% chance of eating the fruits they buy. What is the probability that Kevin eats an apple and Micky does not eat an apple?

Solution:

Given

  • P(Micky buying an apple) = 0.25
  • P(Kevin buying an apple) = 0.4
  • P(eating a fruit that is bought) = 0.6

To Find:P(Kevin eating an apple AND Micky not eating an apple)

Approach

  1. Event-1: Kevin eating an apple
  2. Event-2: Micky not eating an apple
  3. P(Event-1 AND Event-2) = P(Event-1) * P(Event-2)
  4. Event-1: Kevin eating an apple
    1. For Kevin to eat an apple, Kevin has to buy an apple AND eat an apple
    2. P(Event-1) = P(Kevin buying an apple) * P(Kevin eating an apple)
  5. Event-2: Micky not eating an apple
    1. Following cases are possible:
      1. Case-I: Micky buys an apple but does not eat an apple
        1. P(Micky buys an apple but does not eat) = P(Micky buys an apple) * (1- P(Micky eating an apple))
      2. Case-II: Micky does not buy an apple
        1. P(Micky does not buy an apple) = 1- P(Micky buys an apple)
    2. P(Event-2) = P(case-I) + P(case-II)

 

Working Out

  1. Event-1: Kevin eating an apple
    1. P(Event-1) = P(Kevin buying an apple) * P(Kevin eating an apple)
    2. P(Event-1) = 0.4 * 0.6 = 
  2. ​​​Event-2: Micky not eating an apple
    1. Following cases are possible:
      1. Case-I: Micky buys an apple but does not eat an apple OR
        1. P(Micky buys an apple but does not eat) = P(Micky buys an apple) * (1- P(Micky eats an apple)) = 0.25 *(1-0.6) = 0.25 * 0.4=
        2. Case-II: Micky does not buy an apple
        3. P(Micky does not buy an apple) = 1- P(Micky buys an apple) = 1- 0.25 = 0.75 = 3/4​
        4. P(Event-2) = P(case-I) + P(case-II)=​​ P(Event-1 AND Event-2) = P(Event-1) * P(Event-2) = 
QUESTION: 3

What is the probability that when the letters of the word SERENDIPITY are randomly rearranged, the first alphabet of the resulting word is neither T nor E?

Solution:

Given:

  • The word SERENDIPITY
    • Total alphabets = 11
    • {D,E, E, I, I, N, P, R, S, T, Y}
    • 2 Es and 2 Is

To find: The probability that when the letters of the word are randomly arranged, the first alphabet is not T or E

Approach:

  1. P(1st letter not being T or E) = 1 – P(1st letter being T or E)
  2. P(1st letter being T or E) = P(1st letter being T) + P(1st letter being E)
  3. So, to answer the question, we need to find P(1st letter being T) and P(1st letter being E)

Working Out:

  • SERENDIPITY has 2Es and 1T and 11 total alphabets
  • Looking at the answer choices, we see that the correct answer is Option C
QUESTION: 4

In a classroom, 40% of the boys had read a particular book. What was the probability that a student who was randomly selected from the classroom was a girl who had read the book?

(1) Three-eighths of all students in the classroom had read the book

(2) 20 girls in the classroom had not read the book

Solution:

teps 1 & 2: Understand Question and Draw Inferences

Given:

  • We are given information about the students in the classroom based on 2 attributes:
    • Gender – Girls, Boys
    • Book – Read, Not Read

 

  • Accordingly, we can draw the following table to represent the information given in this question:
  •  Let the total number of Boys in the class be B and the total number of girls be G
    • As given: 40% of the boys read. So, their number is equal to 0.4B
    • Therefore, the number of boys who do not read = B – 0.4B = 0.6B
      • (Note: In this solution, the given information will be displayed in black font color in the table, while the inferred information will be displayed in blue font color)

 

 

To find: P(Choosing a girl has read the book)

 

  • Let the number of girls who have read the book be GR
  • So, Required Probability = 

Step 3: Analyze Statement 1 independently

Statement 1 says that ‘Three-eighths of all students in the classroom had read the book’

Therefore, 

 

  • This equation will give us an expression for GR in terms of B and G
  • So, we will get an expression for the Required Probability in terms of B and G. However, since we do not know the values of B and G (or the ratio B/G), we will not be able to find the value of the Required Probability
  • Therefore, Statement 1 alone is not sufficient.

Step 4: Analyze Statement 2 independently

Statement 2 says that ‘20 girls in the classroom had not read the book’

  • Therefore, GR = G – 20
    • (And Total Students who Read = 0.4B + G – 20)
  • So, Required Probability = 
  • However, since we do not know the values of G and B, we cannot find the value of the Required Probability
  • So, Statement 2 alone is not sufficient

 

Step 5: Analyze Both Statements Together (if needed)

Combining the 2 statements, we can write:

  • This is a linear equation with 2 unknowns. So, we cannot find unique values of B and G from this equation.
  • Therefore, we cannot find unique values of the required probability even after combining the 2 statements.

Answer: Option E

 

 

QUESTION: 5

A playgroup is made up entirely of n pairs of siblings, including the siblings Adam and Josh. 4 members of the playgroup are chosen to represent it in a competition. What is the value of n?

(1)  The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5

(2)  The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Given:

  • Playgroup has n sibling pairs
    • So, Total members = 2n
  • Adam-Josh are a sibling pair
  • 4 members selected from the group
    • Number of ways in which 4 members can be selected from 2n members = 2nC4

​​To find: n = ?

Step 3: Analyze Statement 1 independently

  • The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
  • Let’s first find out the expression for the Probability that Adam and Josh are among the 4 members chosen. Then, we’ll equate this expression to 25
  • Calculating the number of ways in which the favorable event can happen
    • Adam and Josh are definitely there among the 4 representatives
    • So, when choosing the 4 representatives, our task simplifies to choosing 2 representatives out of the 2n-2 members that remain after Adam and Josh are taken out
    • The number of ways in which we can chose 2 representatives out of 2n – 2 members =  
  • Calculating the number of total outcomes possible
  • Total number of members = 2n
  • Number of representatives to be chosen = 4
  • The number of ways in which 4 representatives can be chosen out of 2n members = 

Calculating the Probability that the favorable event will happen

⇒n=3 (rejecting the negative value since number of people cannot be negative)

Thus, Statement 1 alone is sufficient to arrive at a unique value of n.

 

Step 4: Analyze Statement 2 independently

  • The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5
  • Let’s first find out the expression for the Probability that 2 sibling pairs are chosen to represent the playgroup. Then, we’ll equate this expression to 1/5
  • Calculating the number of ways in which the favorable event can happen
    • Total number of sibling pairs = n
    • Number of sibling pairs to be chosen = 2
    • The number of ways in which 2 pairs can be chosen out of n pairs =
  • Calculating the number of total outcomes possible
    • Total number of members = 2n
    • Number of representatives to be chosen = 4
    • The number of ways in which 4 representatives can be chosen out of 2n members =
  • Calculating the Probability that the favorable event will happen

⇒n=3

 (rejecting the negative value since number of people cannot be negative)

Thus, Statement 2 alone is sufficient to arrive at a unique value of n.

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already arrived at a unique answer in each of Steps 3 and 4, this step is not required

Answer: Option D

 

QUESTION: 6

If the probability that Brendon, Daniel and Kane score more than or equal to 700 on the GMAT is 0.4, 0.5 and 0.6 respectively, what is the probability that at least 2 of them score less than 700?

Solution:

Given

  • P(Brendon scoring more than or equal to 700) = 0.4
  • P(Daniel scoring more than or equal to 700) = 0.5
  • P(Kane scoring more than or equal to 700) = 0.6

 

To Find: Probability that at-least 2 of them score less than 700?

Approach

  1. P(at-least 2 of them score less than 700) = P(exactly 2 of them score less than 700 or all 3 of them score less than 700)
    1. Event-1: Two of them score less than 700
    2. Event-2: All 3 of them score less than 700
  2. P(Event) = P(Event-1) + P(Event-2)
  3. Event-1: Two of them score less than 700
    1. The possible combinations can be:
      1. Case-I: Brendon and Daniel scoring less than 700 AND kane scoring more than or equal to 700 OR
      2. Case-II: Daniel and Kane scoring less than 700 AND Brendon scoring more than or equal to 700 OR
      3. Case-III: Kane and Brendon scoring less than 700 AND Daniel scoring more than or equal to 700
      4. P(Event-1) = P(case-I) + P(case-II) + P(case-III)
  4. Event-2: All 3 of them scoring less than 700
    1. Event-2 = (Bredon scoring less than 700 AND Daniel scoring less than 700 AND Kane scoring less than 700)
    2. P(Event-2) = P(Brendon scoring less than 700) * P(Daniel scoring less than 700) *P(Kane scoring less than 700)

 

Working Out

  1. Event-1: Two of them scoring less than 700
    1. Case-I: (1-0.4) * (1- 0.5) * 0.6 = 0.18
    2. Case-II:  0.4 *(1-0.5) *(1-0.6)  = 0.08
    3. Case-III: (1-0.4) * 0.5 * (1- 0.6) = 0.12
    4. P(Event-1) = 0.18 + 0.08 + 0.12 =0.38
  2. Event-2: All 3 of them scoring less than 700
    1. P(Event-2) = (1-0.4) * (1-0.5) * (1-0.6) = 0.6 * 0.5 * 0.4 = 0.12
  3. P(Event) = P(Event-1) + P(Event-2) = 0.38 + 0.12 = 0.50
QUESTION: 7

A teacher prepares 20 chits, each chit having a unique integer out of the first 20 positive integers. A student is asked to draw a chit at random. What is the probability that the chit drawn by the student does not have a prime number?

Solution:

Given: 20 chits – each having a unique integer from 1 to 20

To find: P(Choosing a non-prime number)              

Approach:

  1. Since the number of prime numbers between 1 – 20 is lesser than the number of non-prime (composite) numbers between 1 – 20, the non-event method will be easier to use in this question.
    1. So, we can write: P(Choosing a non-prime number) = 1 – P(Choosing a prime number)

2. P(Choosing a prime number) =

3. So, to answer the question, we will determine the number of prime numbers between 1 and 20

Working Out:

  1. The prime numbers between 1 and 20 are: {2, 3, 5, 7, 11, 13, 17, 19}
    1. Upon counting, we see that there are 8 prime numbers between 1 and 20

 

  1. So, P(Choosing a prime number) 
  2. Therefore, P(Choosing a non-prime number) =
QUESTION: 8

The probability that it rains on a certain day in a week is 0.3. What is the probability that it rains on exactly two days in that week?

Solution:
QUESTION: 9

Angelo studied the report by an analyst that predicted the probabilities of the returns four stocks may generate in the next one year. Based on the probabilities of the returns, Angelo invested $100 each in four stocks for a year. He plans to sell a stock at the end of the year if the stock is valued at not less than $140. What is the probability that Angelo sells at least 1 of the stocks?

Solution:

Given

  • Probabilities of stocks generating annual returns
  • $100 invested in each of the four stocks
  • To be sold if value of a stock ≥ 140 at the end of the year

To Find: Probability that at-least 1 of the stock is sold?

Approach

  1. Event: At-least one stock is sold, i.e. at-least one stock has a value ≥ $140, i.e. at-least one stock generates a return ≥ 40%
  2. Non-Event: None of the stocks generate a return of ≥ 40%
    1. Non-Event-1: Stock-1 does not generate a return ≥ 40%
      1. P(Non-Event-1) = 1- P(Stock 1 generating a return ≥ 40%)
    2. Non-Event-2: Stock-2 does not generate a return ≥ 40%
      1. P(Non-Event-2) = 1- P(Stock 2 generating a return ≥ 40%)
    3. Non-Event-3: Stock-3 does not generate a return ≥ 40%
      1. P(Non-Event-3) = 1- P(Stock 3 generating a return ≥ 40%)
    4. Non-Event-4: Stock-4 does not generate a return ≥ 40%
      1. P(Non-Event-4) = 1- P(Stock 4 generating a return ≥ 40%)
    5. P(Non-Event) = P(Non-Event -1) *P(Non-Event-2) *P(Non-Event-3) *P(Non-Event-4)
  3. P(Event) = 1- P(Non-Event)

 

Working Out

  1. Non-Event: None of the stocks generate a return of ≥ 40%
    1. Non-Event-1: Stock-1 does not generate a return ≥ 40%
      1. P(Non-Event-1) = 1- P(Stock 1 generating a return ≥ 40%) = 1- 0.7 = 0.3
    2. Non-Event-2: Stock-2 does not generate a return ≥ 40%
      1. P(Non-Event-2) = 1- P(Stock 2 generating a return ≥ 40%) = 1- 0.3 = 0.7
    3. Non-Event-3: Stock-3 does not generate a return ≥ 40%
      1. P(Non-Event-3) = 1- P(Stock 3 generating a return ≥ 40%) =1- 0.5 = 0.5
    4. Non-Event-4: Stock-4 does not generate a return ≥ 40%
      1. P(Non-Event-4) = 1- P(Stock 4 generating a return ≥ 40%) = 1- 0.8 = 0.2
    5. P(Non-Event) = 0.3 * 0.7 * 0.5 * 0.2 = 0.021
  2. P(Event) = 1- 0.021 = 0.979

 

Answer: C

QUESTION: 10

A group of 30 people includes men, women and children. If one person is to be chosen at random from the group, is the probability that a man is chosen greater than the probability that a woman is chosen?

(1)  The probability that a man is chosen is 50% greater than the probability that a child is chosen.

(2)  The probability that either a woman or a child is chosen is greater than the probability that a man is chosen

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Given:

  • Let number of Men Women, Children be M, W, C respectively.
    • M + W + C = 30
    • Since M, W and C denote the number of people in the group, they must be positive integers

To find: Is P(Choosing a man) > P(Choosing a woman)?

Step 3: Analyze Statement 1 independently

  • The probability that a man is chosen is 50% greater than the probability that a child is chosen.
  • One is tempted to conclude that since this is a linear equation with two unknowns, we’ll not be able to find unique values of M and W, and hence, will not be able to answer the question. However, we should not be so fast in our judgment because we are given a constraint here on the values of M and W:
    • M and W can only be positive integers
  • Let’s evaluate if this constraint, when combined with the above equation, leads us to unique values of M and W or not:
  • We can rewrite the above equation as:
  • Since W must be an integer, M must be a multiple of 3
  • Thus M must be a positive multiple of 3 and must be less than 16

  • Thus, we see that Statement 1 alone is not sufficient to arrive at a unique answer

     

    Step 4: Analyze Statement 2 independently

  • The probability that either a woman or a child is chosen is greater than the probability that a man is chosen
    •    P(Choosing a child) =
  • P(Choosing a Woman or a child) = P(Choosing a Woman) + P(Choosing a child)
  • Therefore, 30–M>M
  •  (Using M + W + C = 30)
  • 30 > 2M
  • So, M < 15
    • If W = 2, M = 1 and C = 27, this condition is satisfied and the answer to the Question is NO
    • If W =8, M = 12 and C = 10, this condition is satisfied and the answer to the Question is YES.
  • So, Statement 2 is not sufficient to find a unique answer to the question.

     

    Step 5: Analyze Both Statements Together (if needed)

  • From Statement 1:

  • From Statement 2: M < 15
  • All the values of M in the table satisfy this inequality
  • Therefore, even after combining both the statements, we don’t’ know if the answer to the question is Yes or No.

    Answer: Option E

QUESTION: 11

If x is an integer such that   what is the probability that 

Solution:

Given

To Find: The probability that  x- 2x -8 =0 ?

We need to find the probability that x = -2, or 4

  • We need to find the probability that x = -2, or 4

Approach

  1. Possible number of ways for which 

2. Possible number of values that x can takeà For finding the values that x can take, we need to solve the inequality 

3. Probability of

4. For solving the inequality, we will use the wavy line method

Woking Out

1. 

  • Now, we can see that the inequality is satisfied at {-3, -2, 0, 1, 2, 3}, i.e. a total of 6 values….(1)
    1. Please note that we have not considered x = -4 as one of the solution points, because for x = -4, the denominator becomes 0
  • As we do not have x = 4 as one of the solutions, x- 2x -8 =0 only when x = -2
  • So, the probability of (x = -2)  = 1/6
  • So, out of the possible 6 values of x,   for only one value of x = -2.
QUESTION: 12

James, Kara and Smith are three friends who have applied for a job at Company X. Their probabilities of getting selected for interview are 2/5,1/3and4/7 respectively. If Company X offers a job to only 25% of the applicants it interviews, what is the probability that only Kara receives a job offer out of the three friends?

Solution:

Given:

  • Selection for Interview
    • P(James selected for interview) = 2/5
  •  
  • P(Kara selected for interview) = 1/3
  •  
  • P(Smith selected for interview) = 4/7
    •  
  • Receiving Job Offer Job after Interview
    • Only 25% of the candidates who are interviewed receive a job offer. This means, only 25% of the candidates who are interviewed clear the interview.
    • So, P(James clears interview) = P(Kara clears interview) = P(Smith clears interview) = 25% = 1/4

To find: Probability that only Kara receives a job offer

Approach:

  1. P(Only Kara receives a job offer) = P(Kara receives offer)*P(James doesn’t receive offer)*P(Smith doesn’t receive offer)
    • = P(Kara receives offer)*[1-P(James receives offer)]*[1 – P(Smith receives offer)]
    • =[P(Kara selected for interview)*P(Kara clears interview)]*[1 – P(James selected for interview)*P(James clears interview)]*[1 – P(Smith selected for interview)*P(Smith clears interview)]
  2. We’re given all the probabilities in the final equation above. So, we can easily find the required probability.

Working Out:

P(Only Kara receives a job offer) = 

Looking at the answer choices, we see that the correct answer is Option B

QUESTION: 13

The probability of a manufacturing company producing a defective item is 0.1. If 5 items are drawn at random from a set of 100 distinct items, what is the probability that at least 2 items would be defective?

Solution:

Given

  • P(defective item) = 0.1
    • P(non-defective item) = 0.9
  • Total number of items = 100
  • Number of items drawn = 5

To Find: Probability that at-least 2 items will be defective

Approach

  1. Event= getting (2, 3, 4, or 5) items defective
  2. Non-Event = getting (0 or 1) item defective
    1. So, we will opt for the non-event method
  3. P(Event) = 1 – P(non-event)
  4. Non- Event: Getting 0 or 1 item defective
  1. Case-I: Getting 0 defective items
  2. Total number of ways in which 0 selected items can be defective = Item-1 is non-defective AND item-2 is non-defective…………AND item-5 is non defective
  3. = P(item-1 being non-defective) * P(item-2 being non-defective) *P(item-3 being non-defective)……….* P(item-5 being non-defective)
  4. Case-II: Getting 1 defective item
  5. P(getting 1 item defective AND 4 items non-defective) = Number of ways of selecting 1 item from 5 items * P(an item being defective) * P(an item being non-defective) *…..P(item being non-defective)
  6. P(Non-Event) = P(case-I) + P(case-II)

 

Working Out

  1. Non-Event: getting 0 or 1 defective items
  1. Case-I: Getting 0 defective items
  2. P(getting 0 defective items) = P(item-1 being non-defective) *P(item-2 being non-defective) …….P(item-5 being non-defective) =0.95…….(1)
  3. Case-II: getting 1 defective item
  4. P(getting 1 item defective AND 4 items non-defective) = Number of ways of selecting 1 item from 5 items * P(an item being defective) * P(an item being non-defective) *…..P(item being non-defective)
  5. P(Non-event-) = 
  6. 2.  P(Event) = 1- P(non-Event) =

 

QUESTION: 14

A box contains balls of two sizes – big and small – and different colors. When one ball is chosen at random from the box, the probability that the ball is blue is 14

. What is the probability that the chosen ball is neither blue nor small?

(1) The probability that the chosen ball is not small is 5/8

(2) The probability that the chosen ball is small but not blue is 9/40

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Given:

  • The balls in a box are defined by 2 attributes:
    • Size – Small or Big
    • Color – Blue or Not Blue
  • P(Blue ball) = 1/4
  • This means, P(Not Blue ball) =   
    • (since there are only 2 possibilities -  a ball is either Blue or Not blue)
  • To find: The probability of choosing a ball that is neither blue nor small

  • Let this probability be X
  • So, we can represent the given information in a matrix form as below:
  • Step 3: Analyze Statement 1 independently

  • The probability that the chosen ball is not small is 58
  • This means, P(Small) =  (since there are only 2 possibilities – a ball is either small or not small)
  • As is clear from the table, we’ve not been able to determine the value of X. So, Statement 1 is not sufficient.

    (Note: You may have marked Statement 1 to be sufficient because you thought that:

  • P(Not Small Not Blue) = P(Not Small)*P(Not Blue)
  • And, since Statement 1 provides P(Not small), this statement was sufficient.
  • The catch here, is that Blue and Small are attributes of the same ball. So, these are not independent attributes.

    The equation P(Not Small Not Blue) = P(Not Small)*P(Not Blue) would have been correct if the question had stated that there were 2 boxes. In one box, the balls are defined by only their color – they are either blue or not blue. In the second box, they are defined by only their size – small or not small. You have to pick one ball from each box. So, what is the probability that you pick a Not Blue ball from the 1st box and a Not Small box from the 2nd box. This is the scenario where the equation mentioned in this note would be applicable)

     

    Step 4: Analyze Statement 2 independently

  • The probability that the chosen ball is small but not blue is 9/40
  •  
  • Let’s represent this information in the matrix:
  • This is a linear Equation with only one unknown.

    So, it is sufficient to find a unique value of X.

     

    Step 5: Analyze Both Statements Together (if needed)

    Since we’ve already arrived at a unique answer in Step 4, this step is not required

     

    Answer: Option B

  •  
QUESTION: 15

The first name and the last name of 5 people are written in two tables above, in a jumbled order. For example, the last name of John may be Garth. Lisa, who doesn’t know the correct first name-last name pair for any of the 5 people in the table, is asked to create 5 first name – last name pairs using each first name and last name in the tables above only once. What is the probability that the pairs she creates includes the correct first name-last name pairs of 2 people in the table?

Solution:

Given:

  • Jumbled-up First Names (henceforth called FN for brevity) and Last Names (LN) of 5 people
  • Lisa makes 5 FN – LN Pairs

To find: Probability that 2 out of the 5 FN – LN pairs made by Lisa are correct

Approach:

  1. P(Event) =  

  1. The favorable event here is to make 2 correct and 3 incorrect FN – LN pairs. This event involves the following tasks:
    • Task 1 – Select the 2 (out of 5) names that Lisa got correct in her list.
    • Task 2 – For the remaining 3 people, whose correct name doesn’t figure in the list created by Lisa, create incorrect FN -LN combinations.
    • Since both these tasks are required for the event, we will multiply the number of ways to do Task 1 and to do Task 2.
    • So, (Number of ways in which 2 correct and 3 incorrect FN – LN pairs can be made)
      • = (Number of ways to select 2 people (out of 5) whose name is correct)*(Number of ways to create incorrect FN – LN combinations for the remaining 3 people)

 

Working Out:

  • Finding total number of ways in which FN – LN combinations may be made
    • For the First name John, Last name choices = 5
    • For the first name Rachel, last name choices left = 4
    • For the first name Mike, last name choices left = 3
    • For the first name Jordan, last name choices left = 2
    • For the first name Sam, last name choices left = 1

 

  • So, total number of FN -LN pairs possible = 5*4*3*2*1 = 120
  • Finding number of ways in which exactly 2 correct FN – LN combinations may be made
    • Number of ways to do Task 1, that is to select the 2 (out of 5) names that Lisa got correct in her list =  Let’s now figure out the number of ways to do Task 2, that is to create incorrect FN -LN combinations for the remaining 3 names
    • First of all, let’s label the 3 First Names and the 3 Last names that remain after Task 1 as under:

We are giving the remaining first and last names these generic labels because we do not know exactly which 3 remains are left behind after Task 1.

  • Total number of FN – LN combinations = 3*2*1 = 6
    • (For FN3, Last Name options are 3. For FN4, only 2 Last Name options will remain. And for FN5, only 1 Last name will remain. So, total number of combinations = 3*2*1)
  • Combinations in which the names of all 3 people are correct = 1
  • Combinations in which the names of 2 people are correct = 0 (because then the left over FN – LN pair must be correct as well. So, either you can get the names of all 3 people correct or of 1 or 0 people correct)
  • Combinations in which the name of 1 person is correct = 1 + 1 + 1 = 3
    • (Explanation: If the name of Person 3 is correct, then the only possible wrong FN-LN combination of persons 4 and 5 is FN4 – LN5 and vice versa. So, there is only 1 way in which the name of Person 3 is correct. Similarly for Person 4 and 5)
  • So, combinations in which the name of no person is correct = 6 – 1 – 3 = 2
  • Thus, we see that the number of ways to do Task 2 = 2
  • So, (Number of ways in which 2 correct and 3 incorrect FN – LN pairs can be made) = 10*2 = 20

 

QUESTION: 16

A mathematics teacher provides the x- and y- coordinates of 10 points in a rectangular coordinate system. These points are (0.5, 4), (1, 4), (2, 4), (3, 4), (3, 5), (4, 5), (5, 5), (6, 5), (6.5, 5) and (7, 5). He asks a student to select 3 points at random. What is the probability that the chosen points form a triangle?

Solution:

Given:

  • A list of 10 points:
    • (0.5, 4), (1, 4), (2, 4), (3, 4)
      • All these 4 points lie on the line y = 4
    • (3, 5), (4, 5), (5, 5), (6, 5), (6.5, 5), (7, 5)
      • All these 6 points lie on the line y = 5
  • 3 points are chosen at random

To find: P(Chosen points form a triangle)

Approach:

  1. P(Chosen points form a triangle) = 1 – P(Chosen points don’t form a triangle)
    • The only way in which the chosen points will not form a triangle is if all the 3 chosen points are collinear
    • P(Chosen points form a triangle) = 1 – P(Chosen points are Collinear)
  2. All the 3 chosen points will be collinear if either all 3 of them lie on the line y = 4 or on the line y = 5
    • So, P(Chosen points are Collinear) = P(Chosen points lie on y = 4) + P(Chosen points lie on y = 5)

Working Out:

  • Finding P(Chosen points lie on y = 4)
    • There are 4 points that lie on line y = 4
    • So, we have to choose 3 collinear points from these 4 points.
    • Number of ways to choose 3 points out of 4 points =

The total number of given points = 10

  • So, the total number of ways in which 3 points may be chosen =

So, P(Chosen points lie on y = 4) =4/120

Finding P(Chosen points lie on y = 5)

  • There are 6 points that lie on line y = 5
  • So, we have to choose 3 points from these 6 points.
    • Number of ways to choose 3 points out of 6 points =

The total number of given points = 10

  • So, the total number of ways in which 3 points may be chosen =

So, P(Chosen points lie on y = 5) = 20/120

  • Getting to the required Probability
    • So, P(Chosen points are Collinear) = 

P(Chosen points form a triangle) =

Looking at the answer choices, we see that the correct answer is Option D

QUESTION: 17

The ratio of men to women in a group is 2:3. 36% of the people in the group do not have a college degree while 60% of the women in the group have a college degree. If 1 person is to be randomly selected from the group, what is the probability that he is a man with a college degree?

Solution:

Given:

  • Men: Women = 2:3
    • This means, 2 out of every 5 people in the group are Men
    • That is, 40% of all people in the group are Men
  • Information is given about the people in the group based on 2 attributes:
    • Gender (Male, Female)
    • Education (College Degree, No College Degree)

 

This information can be represented visually as follows:

 

  • Also, 36% people in the group do not have college degree.
  • 1 person is chosen randomly from the group

To find: Probability that the chosen person is a Man with a College Degree

Approach:

  1. P(Man with a College Degree) = 

  • Where m is the number of men with a college degree
  • And, t is the total number of people in the group

    So, in order to answer the question, we need to find the ratio m/t
    • This ratio represents the percentage of men with a college degree in the group.
       
  • Now, (% of people who are Men with a college degree) = (Total % of people with college degree) – (% of people who are Women with a college degree)
  • (Total % of people with college degree) = 100% - (% of people without college degree) and from the tree structure drawn in the Given section, we can also find the (% of people who are Women with a college degree)
  •  

    Working Out:

  • (Total % of people with college degree) = 100% - 36% = 64%
  • We are given that 60% of all people are Women, and further, 60% of the Women have a college degree

So, (% of people who are Men with a college degree) = 64% - 36% = 28%

  • So, P(Man with a College Degree) = 0.28

Looking at the answer choices, we see that the correct answer is Option D

QUESTION: 18

Frequency Distribution of Integers in Set X and Set Y

 

X and Y are two sets that contain integers as shown the table above. What is the probability that the product of a randomly chosen integer from Set X and a randomly chosen integer from Set Y will be even?

Solution:

Given:

  • Sets X and Y contain integers as shown in the table.
  • 1 integer is chosen from X (the integer chosen from Set X will be referred to as IX from now on) and one from Y (IY)

To find: The probability that IX*IY  is Even

Approach:

  1. P(IX*IY is Even) = 1 – P(IX*IY is Odd)
    • The reason why we are taking the Non-Event Approach here is that there are 3 ways in which the product of 2 integers can be Even ( i. Both integers are even ii. Only IX is even and iii. Only IY is even). However, there is only one way in which IX*IY is odd (when both the integers are odd). So, it’s easier and quicker to solve the question using the Non-Event Approach. 
  2. Since the product IX*IY is odd when IX and IY are odd, we can write:
    • P(IX*IY is Odd) = P(IX is odd)*P(IY is odd)
  3. So, we need to find P(IX is odd) and P(IY is odd)

Working Out:

  • Finding P(IX is odd)

The total number of integers in Set X is 30

  • So, the total number of ways in which one integer can be selected from Set X = 
  • The odd integers in Set X (3, 5, 7) are highlighted in the table. 3 occurs 4 times in Set X, 5 occurs 6 times and 7 occurs 8 times
  • So, the total number of odd integers in Set X = 4 + 6 + 8 = 18
  • So, the number of ways in which one ODD integer can be selected from Set X= 
  • Therefore, P(IX is odd) = 
  • Finding P(IY is odd)
  • The total number of integers in Set Y is 12
  • So, the total number of ways in which one integer can be selected from Set 
  • All the integers in Set Y are odd (highlighted in the table) except the two occurrences of 24.
  • So, the total number of odd integers in Set Y = 12 – 2 = 10
  • So, the number of ways in which one ODD integer can be selected from Set Y =
  • Therefore, P(IY is odd) =  

Finding the Required Probability

Looking at the answer choices, we see that the correct answer is Option D

QUESTION: 19

In the given figure, ABC is an equilateral triangle such that AD= DB and DF is parallel to EG. If a point is chosen at random inside the triangle, what is the probability that the point would lie inside the quadrilateral DEGF?

Solution:

Given

  • ABC is an equilateral triangle
  • DEGF is a quadrilateral with two angles of 90   each
  • AD = DB
  • DF || EG
    • As DF || EG and ∠DEG = ∠FGE = 90  , ∠EDF = ∠DFG = 90 
    • So, DEFG is either a rectangle or a square
  • A point P is selected at random inside the triangle

To Find: Probability of the point P lies inside the quadrilateral DEFG?

Approach

  1. Possible number of ways in which P can lie inside quadrilateral DEFG = Ar(DEFG)
  2. Total number of ways in which P can lie inside triangle ABC = Ar(ABC)
  3. Therefore, Probability of P lying inside DEFG =
  4. As we do not know the values of any of the sides, we will need to express the areas of both quadrilateral DEFG and triangle ABC in a common term
  • For this purpose, let’s assume the length of side of triangle ABC to be x
  • So, Ar(ABC) =
  1. So, we need to express the Ar(DEFG) in terms of x.
    1. As DEFG is a rectangle, we will try to express the sides of DEFG in terms of x
  2. To find the lengths of DEFG, we will focus on triangles ADE and EGC (as they have DE and EG as one of their sides)
    1. We will use the standard trigonometric ratios to express DE and EG in terms of x

Working Out

  • In triangle EGC, we have ∠GCE = 60o and ∠EGC = 90o.
    1. So, that leaves us with ∠CEG = 30o
    2. However, we right now do not know any of the side lengths of this triangle.
  • We know that AC is a straight line. So, ∠AED + ∠DEG + ∠CEG = 180o
    1. That gives us ∠AED = 60o
  • In triangle ADE, we have ∠DAE = ∠AED = 60o
    1. So, ∠ADE = 60o
    2. Thus, ADE is an equilateral triangle with AD = DE = EA
    3. But we know that AD = DB = x/2
  • So, we have DE =x/2….(1)
    1. Hence, we have one side of DEFG in terms of x
  • Coming back to triangle EGC, as we have AE =
  • , we can write EC = 
  • Thus, triangle EGC is a 30o- 60o– 90o triangle with one of the sides EC =
  • 6. Hence probability of point P lying inside DEFG =
QUESTION: 20

For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If one integer is selected at random between 5! And 5! + 10, inclusive, what is the probability that the chosen number will have only two factors?

Solution:

Given:

  • 1 integer is chosen from this list at random

To find: Probability that the chosen integer will have only 2 factors

Approach:

  1. A number that has only 2 factors is a Prime number. So, we need to find the probability of choosing a prime number between 120 and 130, inclusive

2. 

  1. So, to answer the question, we need to count the number of Prime Numbers in the list

Working Out:

  • Finding the number of Prime numbers in the list
    • The square root of all numbers in this list will be less than 12. So, we only need to consider the division of these candidates with prime numbers less than 12. These prime numbers are {2, 3, 5, 7, 11}
      • Note, this is an application of the process to check if a number is prime. If you have any doubt about this, please revise the concept file.
    • Checking for divisibility by 2:  All the even numbers in the list are divisible by 2. So, they are definitely not prime. So, at this stage, only the odd numbers in the list have the possibility of being prime.

 

So, the candidates for primality at this stage are:

{121, 123, 125, 127, 129}

  • Checking for divisibility by 3: In the above list, the numbers 123 (sum of digits = 1 + 2 + 3 = 6) and 129 (sum of digits = 1 + 2 + 9 = 12) are divisible by 3. So, they’re ruled out for the possibility of being prime.

  • Checking for divisibility by 5: The number 125 is divisible by 5. So, it too is ruled out for the possibility of being prime.

  • Checking for divisibility by 7: Neither of the 2 remaining candidates, 121 and 127, is divisible by 7. So, both of them remain possible candidates for being prime numbers.

  • Checking for divisibility by 11: However, 121 is divisible by 11. So, it is ruled out as a candidate for being prime number.

  • The only number standing at the end of all our divisibility checks is 127. So, since this number has passed the test of Primality, we can be confident that 127 is a prime number.

 

  • Thus, there is only one prime number in the given list
  • Finding the Required Probability.

QUESTION: 21

Mary has p pencils and q pens in her bag while Sam has r pencils and s pens in his bag. If Mary and Sam pick up an item at random from their respective bags, who among them has a higher probability of picking up a pencil?

1) p > r

2) q > s

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Step 3: Analyze Statement 1 independently

  1. p > r

It does not tell us:

  • the relation between q and s
  • how much greater is p than r

Insufficient to answer

 

Step 4: Analyze Statement 2 independently

     2.  q > s

It does not tell us:

  • the relation between p and r
  • how much greater is q than s

 

Insufficient to answer

Step 5: Analyze Both Statements Together (if needed)

  1. p > r
  2. q > s

Although it gives us the relation between p and r and q and s, it does not tell us how much greater is p than r and q than s.

For example, consider these two cases:

Insufficient to answer

Answer: E

QUESTION: 22

A box contains orange, green and blue balls. If one ball is chosen at random from the box, what is the probability that the chosen ball is orange?

(1)  The probability that the chosen ball is blue is one-fourth of the probability that the chosen ball is not blue

(2)  If there were 15 fewer orange balls in the box, the probability that the chosen ball is orange would have been equal to the probability that the chosen ball is blue

Solution:

Given:

  • Let the number of orange, green and blue balls be R, G and B respectively.

To find: P(R)

Step 3: Analyze Statement 1 independently

  • The probability that the chosen ball is blue is one-fourth of the probability that the chosen ball is not blue

But the expression for the probability that the chosen ball is blue =BR+G+B=

  •  We do not know the exact values of R and G. Nor do we know the value of the ratio R:G
  • So, we cannot find a unique value of P(R ) from the above equation.​

Statement 1 is not sufficient to find a unique answer to the question

Step 4: Analyze Statement 2 independently

  • If there were 15 fewer orange balls in the box, the probability that the chosen ball is orange would have been equal to the probability that the chosen ball is blue
    • Number of orange balls = R – 15
    • Number of blue balls = B
    • Number of green balls = G
    • So, The total number of balls in this case = G + B + (R – 15)

  • Since we do not know the unique values of R and G, we cannot find the value of P(R)

Thus, Statement 2 alone is not sufficient to answer the question

Step 5: Analyze Both Statements Together (if needed)

  • So, total number of balls,
  • So,
  • Since we don’t know the value of R, we cannot find this probability.
  • Therefore, the 2 statements together are also not sufficient to answer the question.

    Answer: Option E

  •  
QUESTION: 23

Two distinct fair dice are rolled together. If a fair coin and a biased coin are also tossed together, what is the probability of getting 1 head and 1 tail on the coins and the sum of the two dice greater than 6? Assume that the probability of getting a head on the biased coin is 0.75.

Solution:

Given

  • Two distinct dice are rolled together
  • Fair coin is tossed
    • P(head) = P(tail) = 0.5
  • Biased coin is tossed

To Find: Probability of getting 1 head and 1 tail and the sum of the dice > 6

Approach

  1. Event-1: Probability of getting 1 head and 1 tail
  2. Event-2: Probability of getting the sum of two dice > 6
  3. P(Event-1 AND Event-2) = P(Event-1) * P(Event-2)
  4. Event-1: Getting 1 head and 1 tail on tossing of coins
    1. Following cases are possible:
      1. Case-I: Getting a head on the fair coin and a tail on the biased coin OR
      2. Case-II: Getting a tail on the fair coin and a head on the biased coin
      3. P(Event-1) = P(case-I) + P(case-II)
        1. As we know the probability of getting a head and a tail on both the coins, we can calculate the probability of getting a head and a tail on tossing the 2 coins
  5. Event-2: Getting the sum of dice > 6
    1. Event-2: Getting the sum of dice > 6, i.e. {7, 8, 9, 10, 11 or 12}
    2. Non-Event-2: Getting the sum of dice ≤ 6, i.e. {2, 3, 4, 5 or 6}
    3. As the non-event has lesser number of cases to calculate, we will solve this with the non-event method
    4. So, P(Non-Event-2) = P(getting a sum of 2) + P(getting a sum of 3) +……….P(getting a sum of 6)
    5. So, we need to find the number of ways of getting a sum of  2, 3,…..6 and then divide it by the number of possible ways of rolling the two dices.
    6. P(Event-2) = 1- P(Non-Event-2
QUESTION: 24

In the figure above, rectangle PQRS is a shaded region inside the square ABCD. What is the probability that a point chosen at random from the square ABCD will lie inside the shaded rectangle PQRS?

(1) The length of a diagonal of rectangle PQRS is 55% the length of a diagonal of square ABCD

(2) The length of side PQ is 20% greater than the length of side QR

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Given: The figure that shows rectangle PQRS inside square ABCD

To find: Probability that a randomly chosen point from the square lies inside the rectangle PQRS

Step 3: Analyze Statement 1 independently

Statement 1 says that ‘The length of a diagonal of rectangle PQRS is 55% the length of a diagonal of square ABCD’

The above equation has 2 unknowns:

  • So, we cannot find a unique value of the required ratios from this single equation
  • Therefore, Statement 1 is not sufficient.

Step 4: Analyze Statement 2 independently

Statement 2 says that ‘The length of side PQ is 20% greater than the length of side QR’

However, in order to answer the question, we need to know the value of

  • Since Statement 2 doesn’t provide us with these ratios, it is not sufficient.

 

Step 5: Analyze Both Statements Together (if needed)

  • Rejecting the negative root since the ratio of two lengths cannot be negative
  • Thus, by combining the 2 statements, we get a unique value of the ratio QR/AB
  • By substituting the equation from Statement 2, we can also get a unique value of the ratio PQ/AB
  • So, we will be able to find the value of 
  • Thus, the 2 statements together are sufficient to answer the question

 

QUESTION: 25

A basket contains red and green balls. If two balls are drawn from the basket without replacement, what is the probability that both balls are red?

(1) If the balls were drawn with replacement, the probability of both balls being red would have been (2/5)2

(2) The ratio of red and green balls in the basket is 2:3

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

  • Let’s assume the number of red balls be r and green balls be g

Number of ways to draw 2 balls out of r + g balls =  

To Find: Probability of 2 red balls being picked up =

Need to find values of r and g.

Step 3: Analyze Statement 1 independently

(1) If the balls were drawn with replacement, the probability of both balls being red would have been (25)2

 

  • Probability of picking up two red balls with replacement = 

Probability of 2 red balls being picked up =

Insufficient to answer the question.

 

Step 4: Analyze Statement 2 independently

(2) The ratio of red and green balls in the basket is 2:3

Insufficient to answer the question.

 

Step 5: Analyze Both Statements Together (if needed)

Both the statements give us the same information 3r = 2g.

Insufficient to answer the question.

 

Answer: E

 

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