# Modern Math - Probability - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Modern Math - Probability. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2020 QADI | Modern Math - Probability**

Ashok has a bag containing 40 cards, numbered with the integers from 1 to 40. No two cards are numbered with the same integer. Likewise, his sister Shilpa has another bag containing only five cards that are numbered with the integers from 1 to 5, with no integer repeating. Their mother, Latha, randomly draws one card each from Ashok’s and Shilpa’s bags and notes down their respective numbers. If Latha divides the number obtained from Ashok’s bag by the number obtained from Shilpa’s, what is the probability that the remainder will not be greater than 2?

- A.
0.91

- B.
0.87

- C.
0.94

- D.
0.73

- E.
0.8

Answer: Option B

**Explanation** :

The number of ways of selecting one card from Ashok's bag and other from Shilpa’s bag = ^{40}C_{1} × ^{5}C_{1} = 200 ways

Favourable cases:

**Cards 1 or 2 or 3**: If card 1 or 2 or 3 is drawn from Shilpa’s bad remainder in each case will be less than or equal to 2.

∴ Favourable cases = ^{40}C_{1} × ^{3}C_{1} = 120 ways

**Card 4**: If card 4 is drawn from Shilpa’s bag we need to eliminate those cases when the remainder will be 3.

⇒ We need to eliminate those cards from Ashok’s bag which have number of the form 4x + 3.

∴ Cards of the form 4x + 3 are 3, 7, 11, …., 39 i.e., 10 cards.

∴ Favourable cases = (40 – 10) × 1 = 30 ways

**Card 5**: If card 5 is drawn from Shilpa’s bag we need to eliminate those cases when the remainder will be 3 or 4.

⇒ We need to eliminate those cards from Ashok’s bag which have number of the form 5x + 3 or 5x + 4.

∴ Cards of the form 5x + 3 are 3, 8, 13, …., 38 i.e., 8 cards.

∴ Cards of the form 5x + 4 are 4, 9, 14, …., 39 i.e., 8 cards.

∴ Favourable cases = (40 - 16) × 1 = 24 ways

∴ Total Favourable cases = 120 + 30 + 24 = 174 ways

⇒ Probability of remainder not greater than 2 = 174/200 = 0.87

Hence, option (b).

Workspace:

**IIFT 2019 QA | Modern Math - Probability**

According to birth registration data available with the South Delhi Municipal Corporation, 7 babies were born in a particular week in a private hospital. What is the probability that three babies were born on the same day of the week?

- A.
1800/7

^{5} - B.
1600/7

^{5} - C.
2100/7

^{5} - D.
2400/7

^{5}

Answer: Option A

**Explanation** :

The question is wrong and was discarded by IIFT.

Workspace:

**XAT 2018 QADI | Modern Math - Probability**

A coin of radius 3 cm is randomly dropped on a square floor full of square shaped tiles of side 10 cm each. What is the probability that the coin will land completely within a tile? In other words, the coin should not cross the edge of any tile.

- A.
0.91

- B.
0.5

- C.
0.49

- D.
0.36

- E.
0.16

Answer: Option E

**Explanation** :

Let’s consider one of the tiles with side 10 cm. For the coin to land completely within the tile, its centre should fall anywhere inside the square of side 4 cm.

∴ Favourable area = 4 × 4 = 16

Total area available = 10 × 10 = 100

∴ Required probability = 0.16

Hence, option (e).

Workspace:

**XAT 2017 QADI | Modern Math - Probability**

A dice is rolled twice. What is the probability that the number in the second roll will be higher than that in the first?

- A.
5/36

- B.
8/36

- C.
15/36

- D.
21/36

- E.
None of the above

Answer: Option C

**Explanation** :

Total combinations when a die is rolled twice = 6 × 6 = 36

**Case 1:** The first roll = 1

The second roll = (2, 3, 4, 5, 6)

∴ Favourable outcomes = 5

**Case 2**: The first roll = 2

The second roll = (3, 4, 5, 6)

∴ Favourable outcomes = 4

**Case 3**: The first roll = 3

The second roll = (4, 5, 6)

∴ Favourable outcomes = 3

**Case 4**: The first roll = 4

The second roll = (5, 6)

∴ Favourable outcomes = 2

**Case 5**: The first roll = 5

The second roll = (6)

∴ Favourable outcomes = 1

Hence, the number of favourable outcomes = 5 + 4 + 3 + 2 + 1 = 15

Therefore, the probability = 15/36

**Alternately,**

Let us calculate the number of ways a different number comes up both the die = 6 × 5 = 30.

Out of these 30 ways, in half of these first die will have higher number and in other half second die will have higher number.

∴ Probability that second die has higher number = 15/36.

Hence, option (c).

Workspace:

**IIFT 2017 QA | Modern Math - Probability**

Writex Brown, an E-commerce company gives home delivery of its valuable products, after receiving final order in their website, by different modes of transportation like bike, scooter, tempo and truck. The probabilities of using bike, scooter, tempo and truck are respectively 2/9, 1/9, 4/9, and 2/9. The probabilities of delivering the product late to the destination by using these modes of transport are 3/5, 2/5, 1/5, and 4/5. If the product reaches the destination in time, find the probability that the company has used scooter to reach the destination.

- A.
1/10

- B.
4/25

- C.
3/25

- D.
None of these

Answer: Option C

**Explanation** :

This is a case of conditional probability.

The product being delivered on time implies no delay in delivering the product.

P(Scooter | No Delay) = [P(Scooter and No Delay Using Scooter] / [P(No Delay)]

[P(Scooter and No Delay] = P(Scooter) × P(No Delay Using Scooter)

= (1/9) × [1 − (2/5)] = (1/9) × (3/5) = 3/45

P(No delay) = [(2/9) × (2/5)] + [(1/9) × (3/5)] + [(4/9) × (4/5)] + [(2/9) × (1/5)]

= (4/45) + (3/45) + (16/45) + (2/45) = 25/45

∴ Required probability = (3/45) / (25/45) = 3/25

Hence, option 3.

Workspace:

**IIFT 2016 QA | Modern Math - Probability**

The student mess committee of a reputed Engineering College has n members. Let P be the event that the Committee has students of both sexes and let Q be the event that there is at most one female student in the committee. Assuming that each committee member has probability 0.5 of being female, the value of n for which the events A and B are independent is

- A.
2

- B.
3

- C.
4

- D.
None of the above

Answer: Option B

**Explanation** :

P = event that both males and females are present in the committee = 1/n + 2/n +....+ (n – 1)/n

= [1 + 2 + 3 + … + (n – 1)]/n

= [(n)(n − 1)/2n] = (n – 1)/2

Q = probability that at most one female is present in the committee = 1/n

Probability of P⋂Q = P × Q

P ⋂ Q = 1/n as “only one female in the committee is the intersection of the two events”.

∴ (1/n) = [(n − 1)/2] × (1/n)

∴ n – 1 = 2 i.e. n = 3

Hence, option 2.

Workspace:

**IIFT 2015 QA | Modern Math - Probability**

The internal evaluation for Economics course in an Engineering programme is based on the score of four quizzes. Rahul has secured 70, 90 and 80 in the first three quizzes. The fourth quiz has ten True-False type questions, each carrying 10 marks. What is the probability that Rahul’s average internal marks for the Economics course is more than 80, given that he decides to guess randomly on the final quiz?

- A.
12/1024

- B.
11/1024

- C.
11/256

- D.
12/256

Answer: Option B

**Explanation** :

Average marks of the first three quizzes = 80

So, for Rahul to have average internal marks more than 80, he has to score more than 80 marks in the last quiz.

This is possible if he attempts 10 questions or 9 questions correctly.

Number of ways this can be done = 1 + 10C9 = 1 + 10 = 11

Total number of ways the quiz can be solved = 210 = 1024

∴ The required probability = 11/1024

Hence, option 2.

Workspace:

**IIFT 2015 QA | Modern Math - Probability**

In a reputed engineering college in Delhi, students are evaluated based on trimesters. The probability that an Engineering student fails in the first trimester is 0.08. If he does not fail in the first trimester, the probability that he is promoted to the second year is 0.87. The probability that the student will complete the first year in the Engineering College is approximately:

- A.
0.8

- B.
0.6

- C.
0.4

- D.
0.7

Answer: Option A

**Explanation** :

We consider that the student fails in the first year if he fails in the first trimester.

Therefore, the probability that the student will complete the first year the first year in the Engineering College is approximately = Probability that he passes 1st trimester × Probability that he passes 1st semester and is promoted to the second year) = 0.92 × 0.87 ≈ 0.8

Hence, option 1.

Workspace:

**XAT 2015 QA | Modern Math - Probability**

Ramesh plans to order a birthday gift for his friend from an online retailer. However, the birthday coincides with the festival season during which there is a huge demand for buying online goods and hence deliveries are often delayed. He estimates that the probability of receiving the gift, in time, from the retailers A, B, C and D would be 0.6, 0.8, 0.9 and 0.5 respectively.

Playing safe, he orders from all four retailers simultaneously. What would be the probability that his friend would receive the gift in time?

- A.
0.004

- B.
0.006

- C.
0.216

- D.
0.994

- E.
0.996

Answer: Option E

**Explanation** :

Required Probability = 1 – P(receiving no gift)

P(receiving no gift) = 0.4 × 0.2 × 0.1 × 0.5

= 0.0040

1 – P(receiving no gift) = 0.996

Hence, option 5.

Workspace:

**IIFT 2013 QA | Modern Math - Probability**

Suppose there are 4 bags. Bag 1 contains 1 black and *a*^{2} – 6*a* + 9 red balls, bag 2 contains 3 black and *a*^{2} – 6*a* + 7 red balls, bag 3 contains 5 black and *a*^{2} – 6*a* + 5 red balls and bag 4 contains 7 black and *a*^{2} – 6*a* + 3 red balls. A ball is drawn at random from a randomly chosen bag. The maximum value of probability that the selected ball is black, is

- A.
$\frac{16}{{a}^{2}-6a+10}$

- B.
$\frac{20}{{a}^{2}-6a+10}$

- C.
$\frac{1}{16}$

- D.
None of the above

Answer: Option D

**Explanation** :

Each bag has *a*^{2} – 6*a* + 10 balls.

Bags 1, 2, 3 and 4 contain 1, 3, 5 and 7 black balls respectively.

Probability of selecting a black ball from a specific bag is

$\frac{n}{{a}^{2}-6a+9}$

where n is the number of black balls in that bag.

A bag is selected at random.

∴ Probability of selecting a particular bag = $\frac{1}{4}$

∴ Probability that the ball selected from that randomly chosen bag is black

= $\frac{1}{4}$$\left(\frac{1}{{a}^{2}-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{3}{{a}^{2}-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{5}{{a}^{2}-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{7}{{a}^{2}-6a+10}\right)$

=$\frac{1}{4}$$\left(\frac{16}{{a}^{2}-6a+10}\right)$ = $\left(\frac{4}{{a}^{2}-6a+10}\right)$

Hence, option 4.

Workspace:

**IIFT 2012 QA | Modern Math - Probability**

Ashish is studying late into the night and is hungry. He opens his mother’s snack cupboard without switching on the lights, knowing that his mother has kept 10 packets of chips and biscuits in the cupboard. He pulls out 3 packets from the cupboard, and all of them turn out to be chips. What is the probability that the snack cupboard contains 1 packet of biscuits and 9 packets of chips?

- A.
$\frac{6}{55}$

- B.
$\frac{12}{73}$

- C.
$\frac{14}{55}$

- D.
$\frac{7}{50}$

Answer: Option C

**Explanation** :

There are at least three packets of chips in the cupboard. There are 10 packets in all.

∴ (Number of packets of chips, Number of packets of biscuits) ≡ (3, 7) or (4, 6) or (5, 5) or (6, 4) or (7, 3) or (8, 2) or (9, 1) or (10, 0)

The number of ways in which three packets of chips can be drawn ^{3}C_{3} + ^{4}C_{3} + ^{5}C_{3} + ^{6}C_{3} + ^{7}C_{3} + ^{8}C_{3} + ^{9}C_{3} + ^{10}C_{3} = 330

The number of ways in which three packets of chips can be drawn when there are 9 packets of chips = ^{9}C_{3} = 84

∴ Required probability = $\frac{84}{330}$ = $\frac{14}{55}$

Hence, option 3.

Workspace:

**IIFT 2012 QA | Modern Math - Probability**

A student is required to answer 6 out of 10 questions in an examination. The questions are divided into two groups, each containing 5 questions. She is not allowed to attempt more than 4 questions from each group. The number of different ways in which the student can choose the 6 questions is

- A.
100

- B.
160

- C.
200

- D.
280

Answer: Option C

**Explanation** :

As there are 5 questions in each section we have three different cases to be considered.

**Case 1:**

3 questions from section 1 and 3 questions from section 2

Number of ways = ^{5}C_{3 }× ^{5}C_{3} = 100

**Case 2:**

4 questions from section 1 and 2 questions from section 2

Number of ways = ^{5}C_{4} × ^{5}C_{2} = 50

**Case 3:**

2 questions from section 1 and 4 questions from section 2

Number of ways = ^{5}C_{2} × ^{5}C_{4} = 50

∴ Total number of ways = 100 + 50 + 50 = 200

Hence, option 3.

Workspace:

**IIFT 2012 QA | Modern Math - Probability**

The answer sheets of 5 engineering students can be checked by any one of 9 professors. What is the probability that all the 5 answer sheets are checked by exactly 2 professors?

- A.
$\frac{20}{2187}$

- B.
$\frac{40}{2187}$

- C.
$\frac{40}{729}$

- D.
None of the above

Answer: Option B

**Explanation** :

The paper of each student can go to any of the nine professors.

As there are 5 students, there are 9 × 9 × 9 × 9 × 9 = 9^{5} ways in which the papers can be checked by the professors.

Now, number of ways of selecting two professors = ^{9}C_{2}

The five papers can be checked by the two professors in 2^{5} ways, but this will contain two ways in which the papers are checked by just one professor.

∴ The number of ways in which 5 answer sheets are checked by exactly two professors = ^{9}C_{2} × (2^{5} – 2)

Number of ways in which each paper can be checked by a professor = 2

∴ Number of ways such that five papers can be checked by those 2 professors = ^{9}C_{2} × (2^{5} – 2)

∴ Probability = $\frac{{}^{9}{C}_{2}\times ({2}^{5}-2)}{{9}^{5}}$ = $\frac{40}{2187}$

Hence, option 2.

Workspace:

**IIFT 2012 QA | Modern Math - Probability**

The probability that in a household LPG will last 60 days or more is 0.8 and that it will last at most 90 days is 0.6. The probability that the LPG will last 60 to 90 days is

- A.
0.40

- B.
0.50

- C.
0.75

- D.
None of the above

Answer: Option A

**Explanation** :

Probability that the LPG will last ≥ 60 days = 0.8

∴ Probability that the LPG will last < 60 days = 1 – 0.8 = 0.2

Probability that the LPG will last ≤ 90 days = 0.6

∴ The probability that the LPG will last ≥ 60 days and ≤ 90 days will be = (Probability that a LPG will last ≤ 90 days ) – (Probability that a LPG will last < 60 days)

= 0.6 – 0.2 = 0.4

Hence, option 1.

Workspace:

**IIFT 2011 QA | Modern Math - Probability**

A bag contains 8 red and 6 blue balls. If 5 balls are drawn at random, what is the probability that 3 of them are red and 2 are blue?

- A.
80/143

- B.
50/143

- C.
75/143

- D.
None of the above

Answer: Option D

**Explanation** :

5 balls can be drawn in ^{14}C_{5} ways.

Out of these 5 balls, 3 will be red in ^{8}C_{3} × ^{6}C_{2} ways

∴ Required probability = $\frac{{}^{8}{C}_{3}\times {}^{6}{C}_{2}}{{}^{14}{C}_{5}}$ = $\frac{60}{143}$

Hence, option 4.

Workspace:

**IIFT 2009 QA | Modern Math - Probability**

A card is drawn at random from a well shuffled pack of 52 cards.

X: The card drawn is black or a king.

Y: The card drawn is a club or a heart or a jack.

Z: The card drawn is an ace or a diamond or a queen.

Then which of the following is correct?

- A.
P(X) > P(Y) > P(Z)

- B.
P(X) >= P(Y) = P(Z)

- C.
P(X) = P(Y) > P(Z)

- D.
P(X) = P(Y) = P(Z)

Answer: Option C

**Explanation** :

X, Y and Z are events such that

X is the event that the card drawn is black or a king.

Y is the event that the card drawn is a club or a heart or a jack.

Z is the event that the card drawn is an ace or a diamond or a queen.

Consider X.

A black card can be drawn in 26 ways. A king can be drawn in 4 ways. A black king can be drawn in 2 ways.

∴ P(X) = $\frac{26}{52}$ + $\frac{4}{52}$ - $\frac{2}{52}$

∴ P(X) = $\frac{28}{52}$

Consider Y.

A club can be drawn in 13 ways. A heart can be drawn in 13 ways. A jack can be drawn in 4 ways. A Jack of hearts can be drawn in 1 way and a jack of club can be drawn in 1 way.

∴ P(X) = $\frac{13}{52}$ + $\frac{13}{52}$ + $\frac{4}{52}$ - $\frac{1}{52}$ - $\frac{1}{52}$

∴ P(Y) = $\frac{28}{52}$

Consider Z.

An ace can be drawn in 4 ways. A queen can be drawn in 4 ways and a diamond can be drawn in 13 ways. Queen of Diamonds can be drawn in 1 way and ace of diamonds can be drawn in 1 way.

∴ P(Z) = $\frac{4}{52}$ + $\frac{4}{52}$ + $\frac{13}{52}$ - $\frac{1}{52}$ - $\frac{1}{52}$

∴ P(Z) = $\frac{19}{52}$

∴ P(X) = P(Y) > P(Z)

Hence, option 3.

Workspace:

**IIFT 2009 QA | Modern Math - Probability**

Mr. Raheja, the president of Alpha Ltd., a construction company, is studying his company’s chances of being awarded a Rs. 1000 crore bridge building contract in Delhi. In this process, two events interest him. First, Alpha’s major competitor Gamma Ltd, is trying to import the latest bridge building technology from Europe, which it hopes to get before the deadline of the award of contact. Second, there are rumors that Delhi Government is investigating all recent contractors and Alpha Ltd is one of those contractors, while Gamma Ltd is not one of those. If Gamma is able to import the technology and there is no investigation by the Government, then Alpha’s chance of getting contract is 0.67. If there is investigation and Gamma Ltd is unable to import the technology in time, the Alpha’s chance is 0.72. If both events occur, then Alpha’s chance of getting the contract is 0.58 and if none events occur, its chances are 0.85. Raheja knows that the chance of Gamma Ltd being able to complete the import of technology before the award date is 0.80. How low must the probability of investigation be, so that the probability of the contract being awarded to Alpha Ltd is atleast 0.65? (Assume that occurrence of investigation and Gamma’s completion of import in time is independent to each other.)

- A.
0.44

- B.
0.57

- C.
0.63

- D.
0.55

Answer: Option B

**Explanation** :

Let, the probability of government investigation be p.

There are 4 cases:

∴ Probability of contract being awarded to Alpha = [0.8 × (1 ( p) × 0.67] + [0.2 × p × 0.72] + [0.8 × p × 0.58] + [0.2 × (1 ( p) × 0.85] ≥ 0.65

∴ [0.536 × (1 ( p)] + [0.144 × p] + [0.464 × p] + [0.17 × (1 ( p)] ≥ 0.65

∴ p[0.144 + 0.464 ( 0.536 ( 0.17] ≥ 0.65 ( 0.536 ( 0.17

∴ p[(0.098] ≥ (0.056

∴ p[0.098] ≤ 0.056

∴ p ≤ 0.056/0.098

∴ p ≤ 0.5714

Hence, option 2.

Workspace:

**IIFT 2009 QA | Modern Math - Probability**

A doctor has decided to prescribe two new drugs D1and D2 to 200 heart patients such that 50 get drug D1, 50 get drug D2 and 100 get both. The 200 patients are chosen so that each had 80% chance of having a heart attack if given neither of the drugs. Drug D1 reduces the probability of a heart attack by 35 %, while drug D2 reduces the probability by 20%. The two drugs when taken together, work independently. If a patient, selected randomly from the chosen 200 patients, has a heart attack then the probability that the selected patient was given both the drug is:

- A.
0.42

- B.
0.49

- C.
0.56

- D.
0.40

Answer: Option A

**Explanation** :

Probability that patients who have been prescribed only drug D1 have a heart attack = 0.8 × 0.65 = 0.52

∴ 0.52 × 50 = 26 of these will have a heart attack.

Probability that patients who have been prescribed only drug D2 have a heart attack = 0.8 × 0.8 = 0.64

∴ 0.64 × 50 = 32 of these will have a heart attack.

Probability that patients who have been prescribed both drugs D1 and D2 have a heart attack = 0.8 × 0.65 × 0.8 = 0.416

∴ 0.416 × 100 = 41.6 of these will have a heart attack.

∴ Required Probability = 41.6/(26 + 32 + 41.6) = 0.417 ≈ 0.42

Hence, option 1.

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**