Exam - 64 Math: Topic: Probability, Permutation and Combination

Question: Two fair dice are thrown simultaneously. What is the probability that the product of the numbers on the top faces is a perfect square?

Solution:
When two dice are thrown simultaneously, the total number of possible ordered outcomes is
6 × 6 = 36

The product is a perfect square for the following outcomes:
(1, 1), (1, 4), (4, 1), (4, 4), (2, 2), (3, 3), (5, 5), and (6, 6)
Therefore, the number of favourable outcomes is 8.

∴ Required probability = Number of favourable outcomes/Total number of outcomes
= 8/36
= 2/9

Therefore, the probability that the product is a perfect square is 2/9.

Correct answer: খ) 2/9

Question: In how many different ways can the letters of the word 'COFFEE' be arranged?

Solution:
The word 'COFFEE' contains 6 letters.
Here, E occurs 2 times, and F occurs 2 times.

Number of distinct arrangements = 6!/(2! × 2!)
= 720/(2 × 2)
= 180

Therefore, the letters of 'COFFEE' can be arranged in 180 different ways.

Correct answer: ক) 180

Question: A bag contains 7 blue and 5 yellow marbles. If 2 marbles are drawn randomly one after the other without replacement, what is the probability that the first marble is blue and the second is yellow?

Solution:
Total number of marbles = 7 + 5
= 12

Probability that the first marble is blue = 7/12

After drawing one blue marble, 11 marbles remain, of which 5 are yellow.
Probability that the second marble is yellow = 5/11

Therefore, the required probability = (7/12) × (5/11)
= 35/132

Therefore, the probability that the first marble is blue and the second is yellow is 35/132.

Correct answer: ঘ) 35/132

Question: In how many ways can a person choose a complete outfit if they have 5 shirts, 4 trousers, and 3 pairs of shoes available?

Solution:
To form a complete outfit, the person must choose:
1 shirt from 5 shirts,
1 pair of trousers from 4 pairs of trousers, and
1 pair of shoes from 3 pairs of shoes.

∴ Total number of possible outfits = 5 × 4 × 3
= 60

Therefore, the person can choose a complete outfit in 60 different ways.

Correct answer: গ) 60

Question: The probability that A can solve a problem is 2/3, and the probability that B can solve it is 3/4. If both try independently, what is the probability that the problem gets solved?

Solution:
Probability that A cannot solve the problem = 1 - (2/3)
= 1/3
And Probability that B cannot solve the problem = 1 - (3/4)
= 1/4

Since A and B try independently, the probability that neither of them solves the problem is,
= (1/3) × (1/4)
= 1/12

Therefore, the probability that the problem gets solved by at least one of them is
= 1 - (1/12)
= 11/12

Therefore, the probability that the problem gets solved is 11/12.

Correct answer: ক) 11/12

Question: A cricket squad has 15 players. In how many ways can a final playing team of 11 players be selected?

Solution:
Since the order of selection does not matter, we use combinations.

Number of ways to select 11 players from 15 players = ¹⁵C₁₁
= 15!/{11! × (15 - 11)!}
= 15!/(11! × 4!)
= (15 × 14 × 13 × 12 × 11!)/(4 × 3 × 2 × 1 × 11!)
= 15 × 13 × 7
= 1,365

Therefore, the final playing team can be selected in 1,365 different ways.

Correct answer: ঘ) 1,365

Question: If the letters of the word ‘LOGIC’ are arranged at random, what is the probability that the two vowels are adjacent?

Solution:
The word 'LOGIC' contains 5 distinct letters.
Therefore, the total number of arrangements = 5!
= 120

The vowels in the word are O and I.
For the vowels to be adjacent, consider O and I as a single unit.

Thus, the objects to be arranged are: (OI), L, G, and C
These 4 objects can be arranged in = 4! ways

And the two vowels can also be arranged within their unit in = 2! ways

Therefore, the number of favourable arrangements = 4! × 2!
= 24 × 2
= 48

Required probability = Number of favourable arrangements/Total number of arrangements
= 48/120
= 2/5

Therefore, the probability that the vowels occupy adjacent places is 2/5.

Correct answer: গ) 2/5

Question: There are 12 people at a business meeting. If every person shakes hands with every other person exactly once, how many handshakes take place?

Solution:
প্রতিটি করমর্দনে 2 জন ব্যক্তি অংশগ্রহণ করেন।
তাই 12 জন ব্যক্তির মধ্য থেকে যেকোনো 2 জনকে নির্বাচন করার মোট উপায়ই হবে করমর্দনের সংখ্যা।

∴ মোট করমর্দনের সংখ্যা = ¹²C₂
= 12!/{2! × (12 - 2)!}
= 12!/(2 × 10!)
= (12 × 11)/(2 × 1)
= 66

Correct answer: ঘ) 66

Question: A number is chosen at random from the integers 1 to 100 inclusive. What is the probability that it is divisible by 6 or 8 or both?

Solution:
Number of integers from 1 to 100 divisible by 6,
= ⌊100/6​⌋= 16
And number of integers from 1 to 100 divisible by 8,
= ⌊100/8⌋ = 12

The least common multiple of 6 and 8 is, LCM(6, 8) = 24

Therefore, the number of integers divisible by both 6 and 8,
= ⌊100/24⌋ = 4

Using the inclusion-exclusion principle,
Number of integers divisible by 6 or 8 = 16 + 12 - 4
= 24

∴ Required probability = Number of favourable outcomes/Total number of outcomes
= 24/100
= 6/25

Therefore, the required probability is 6/25.

Correct answer: ক) 6/25

Question: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if no digit can be repeated?

Solution:
3 অঙ্কের একটি সংখ্যা গঠনের জন্য শতকের, দশকের ও এককের স্থানে একটি করে অঙ্ক বসাতে হবে।

এখন, শতকের স্থানে 5টি অঙ্কের যেকোনো একটি বসানো যাবে।
একটি অঙ্ক ব্যবহৃত হওয়ার পর দশকের স্থানে অবশিষ্ট 4টি অঙ্কের যেকোনো একটি বসানো যাবে।
এরপর এককের স্থানে অবশিষ্ট 3টি অঙ্কের যেকোনো একটি বসানো যাবে।

∴ মোট সংখ্যা = 5 × 4 × 3
= 60

অথবা,
মোট সংখ্যা = ⁵P₃
= 5!/(5 - 3)!
= 5!/2!
= 60

অতএব, কোনো অঙ্কের পুনরাবৃত্তি না করে মোট 60টি 3 অঙ্কের সংখ্যা গঠন করা যাবে।

Correct answer: খ) 60

Question: In a group of 100 students, 60 study Mathematics, 45 study Physics, and 24 study both subjects. What is the probability that a randomly selected student studies at least one of the subjects?

Solution:
Let M be the set of students who study Mathematics and P be the set of students who study Physics.

Number of students who study at least one subject = n(M ∪ P)

We know,
n(M ∪ P) = n(M) + n(P) - n(M ∩ P)
= 60 + 45 - 24
= 81

Therefore, the required probability = Number of students who study at least one subject/Total number of students
= 81/100
= 0.81

Therefore, the probability that a randomly selected student studies at least one of the subjects is 0.81.

Correct answer: গ) 0.81

Question: A question paper consists of two sections, A and B, each containing 5 questions. A candidate is required to attempt 5 questions in total, selecting at least 2 questions from each section. In how many ways can the candidate make the selection?

Solution:
পরীক্ষার্থীকে মোট 5টি প্রশ্ন নির্বাচন করতে হবে এবং প্রতিটি বিভাগ থেকে কমপক্ষে 2টি প্রশ্ন নিতে হবে।

তাই প্রশ্ন নির্বাচনের সম্ভাব্য বণ্টন দুটি,
A বিভাগ থেকে 2টি এবং B বিভাগ থেকে 3টি প্রশ্ন,
অথবা A বিভাগ থেকে 3টি এবং B বিভাগ থেকে 2টি প্রশ্ন।

প্রথম ক্ষেত্রে নির্বাচনের উপায়সংখ্যা = ⁵C₂ × ⁵C₃
= 10 × 10
= 100

এবং দ্বিতীয় ক্ষেত্রে নির্বাচনের উপায়সংখ্যা = ⁵C₃ × ⁵C₂
= 10 × 10
= 100

সুতরাং, মোট নির্বাচনের উপায়সংখ্যা = 100 + 100
= 200

অতএব, পরীক্ষার্থী মোট 200 উপায়ে প্রশ্ন নির্বাচন করতে পারবে।

Correct answer: ক) 200

Question: Two cards are drawn at random without replacement from a standard pack of 52 cards. What is the probability of getting one spade and one diamond?

Solution:
A standard pack contains 13 spades and 13 diamonds.

The required event can occur in two ways,
1. First card is a spade and the second card is a diamond.
2. First card is a diamond and the second card is a spade.

Therefore, the required probability = (13/52) × (13/51) + (13/52) × (13/51)
= 2 × (13/52) × (13/51)
= 13/102

Therefore, the probability of drawing one spade and one diamond is 13/102.

Correct answer: ঘ) 13/102

Question: There are 10 points in a plane, exactly 4 of which are collinear, and no other set of 3 points is collinear. How many distinct triangles can be formed by joining these points?

Solution:
10টি বিন্দুর মধ্য থেকে যেকোনো 3টি বিন্দু নির্বাচন করলে সম্ভাব্য নির্বাচনের সংখ্যা = ¹⁰C₃
= (10 × 9× 8)/(3 × 2 × 1)
= 120

কিন্তু একই সরলরেখায় অবস্থিত 3টি বিন্দু দিয়ে কোনো ত্রিভুজ গঠন করা যায় না।
একই সরলরেখায় থাকা 4টি বিন্দু থেকে 3টি বিন্দু নির্বাচনের সংখ্যা = ⁴C₃
= 4

সুতরাং, গঠিত ভিন্ন ত্রিভুজের সংখ্যা = ¹⁰C₃ - ⁴C₃
= 120 - 4
= 116

অতএব, মোট 116টি ভিন্ন ত্রিভুজ গঠন করা যাবে।

Correct answer: ক) 116

Question: A bag contains only red, green, and white balls. The probability of selecting a red ball at random is 1/3, and the probability of selecting a white ball at random is 1/2. If the bag contains 9 green balls, what is the total number of balls in the bag?

Solution:
Since the bag contains only red, green, and white balls,

∴ Probability of selecting a green ball = 1 - Probability of selecting a red ball - Probability of selecting a white ball
= 1 - (1/3) - (1/2)
= 1 - (2/6) - (3/6)
= 1/6

Let the total number of balls be x.
Therefore, (1/6) × x = 9
⇒ x = 9 × 6
⇒ x = 54

Therefore, the total number of balls in the bag is 54.

Correct answer: খ) 54

Question: In how many ways can 7 people be seated around a circular table?

Solution:
আমরা জানি,
n জন ব্যক্তিকে একটি গোলটেবিলের চারপাশে বসানোর উপায়সংখ্যা = (n - 1)!

সুতরাং, 7 জন ব্যক্তিকে বসানোর উপায়সংখ্যা = (7 - 1)!
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

অতএব, 7 জন ব্যক্তিকে একটি গোলটেবিলের চারপাশে 720 উপায়ে বসানো যাবে।

Correct answer: গ) 720

Question: A fair die is rolled twice. What is the probability of getting a composite number on the first roll and a prime number on the second roll?

Solution:
The possible outcomes on a fair die are:
1, 2, 3, 4, 5, and 6

The composite numbers are 4 and 6.

Therefore, the probability of getting a composite number on the first roll = 2/6
= 1/3

The prime numbers are 2, 3, and 5.
Therefore, the probability of getting a prime number on the second roll = 3/6
= 1/2

Since the two rolls are independent,
∴ Required probability = (1/3) × (1/2)
= 1/6

Therefore, the required probability is 1/6.

Correct answer: ঘ) 1/6

Question: In how many ways can the letters of the word 'MONDAY' be arranged so that the vowels are together? 

Solution:
'MONDAY' শব্দটিতে মোট 6টি ভিন্ন অক্ষর রয়েছে।
এখানে স্বরবর্ণ দুটি হলো O এবং A।

স্বরবর্ণ দুটি পাশাপাশি রাখতে O এবং A-কে একটি একক বা ব্লক হিসেবে বিবেচনা করি।

তাহলে সাজানোর উপাদানগুলো হলো: (OA), M, N, D এবং Y
অর্থাৎ মোট 5টি উপাদানকে সাজানো যাবে = 5! উপায়ে

আবার, 
O এবং A নিজেদের মধ্যে সাজানো যাবে = 2! উপায়ে

সুতরাং, মোট বিন্যাসের সংখ্যা = 5! × 2!
= 120 × 2
= 240

অতএব, স্বরবর্ণ দুটি পাশাপাশি রেখে 'MONDAY' শব্দটির অক্ষরগুলো 240 উপায়ে সাজানো যাবে।

Correct answer: ক) 240

Question: A bag contains only red, yellow, and green lollipops. What is the probability of selecting a blue lollipop?

Solution:
The bag contains only red, yellow, and green lollipops.
Therefore, the number of blue lollipops in the bag is 0.

∴ Probability of selecting a blue lollipop = Number of blue lollipops/Total number of lollipops
= 0/Total number of lollipops
= 0

Therefore, the probability of selecting a blue lollipop is 0.

Correct answer: ক) 0

Question: If ⁹P_r = 3,024 and ⁹C_r = 126, what is the value of r?

Solution:
আমরা জানি,
ⁿP_r = ⁿC_r × r!
⇒ r! = 3,024/126
⇒ r! = 24
⇒ r! = 4! ; [4! = 4 × 3 × 2 × 1 = 24] 
∴ r = 4 

অতএব, r = 4।

Correct Answer: ঘ) 4

Question: In a lottery, there are 10 prize-winning tickets and 25 blank tickets. If one ticket is drawn at random, what is the probability of getting a prize?

Solution:
Number of prize-winning tickets = 10

∴ Total number of tickets = 10 + 25
= 35

Required probability = Number of prize-winning tickets/Total number of tickets
= 10/35
= 2/7

Therefore, the probability of getting a prize is 2/7.

Correct answer: খ) 2/7

Question: In how many ways can 6 distinct parcels be placed in 3 distinct lockers if each parcel can be placed in any one locker and any locker may be left empty?

Solution:
প্রতিটি পার্সেল 3টি লকারের যেকোনো একটিতে রাখা যাবে।

সুতরাং, 6টি স্বতন্ত্র পার্সেল রাখার মোট উপায়সংখ্যা = 3⁶
= 729

অতএব, পার্সেলগুলো 729 উপায়ে রাখা যাবে।

Correct Answer: ক) 729

Question: A card is drawn at random from a standard pack of 52 cards. What is the probability of getting the queen of clubs or the king of hearts?

Solution:
A standard pack contains exactly one queen of clubs and one king of hearts.
Number of favourable outcomes = 1 + 1
= 2

∴ Total number of possible outcomes = 52

Required probability = Number of favourable outcomes/Total number of outcomes
= 2/52
= 1/26

Therefore, the required probability is 1/26.

Correct answer: ঘ) 1/26

Question: A committee of 5 members is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee must contain exactly 2 ladies?

Solution:
কমিটিতে মোট সদস্য থাকবে 5 জন এবং ঠিক 2 জন মহিলা থাকতে হবে।

অতএব, পুরুষ সদস্যের সংখ্যা = 5 - 2
= 3

4 জন মহিলার মধ্য থেকে 2 জন নির্বাচন করার উপায়সংখ্যা = ⁴C₂
= (4 × 3)/(2 × 1)
= 6

6 জন ভদ্রলোকের মধ্য থেকে 3 জন নির্বাচন করার উপায়সংখ্যা = ⁶C₃
= (6 × 5 × 4)/(3 × 2 × 1)
= 20

সুতরাং, গুণনের মৌলিক নীতি অনুযায়ী মোট উপায়সংখ্যা = ⁴C₂ × ⁶C₃
= 6 × 20
= 120

অতএব, কমিটিটি মোট 120 উপায়ে গঠন করা যাবে।

Correct Answer: ক) 120

Question: How many times as many distinct arrangements can be formed from the letters of ‘READER’ as from the letters of ‘ERROR’?

Solution:
The word 'READER' contains 6 letters.
Here, R occurs 2 times and E occurs 2 times.

Therefore, the number of distinct arrangements of 'READER',
= 6!/(2! × 2!)
= 720/(2 × 2)
= 180

And the word 'ERROR' contains 5 letters.
Here, R occurs 3 times.

Therefore, the number of distinct arrangements of 'ERROR',
= 5!/3!
= 120/6
= 20

∴ Required ratio = Number of arrangements of 'READER'/Number of arrangements of 'ERROR'
= 180/20
= 9

Therefore, the number of arrangements of 'READER' is 9 times the number of arrangements of 'ERROR'.

Correct answer: ঘ) 9 times