# 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:

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