IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

Question: Three unbiased coins are tossed. What is the probability of getting at least 2 tails?

Solution:
Total outcomes = {TTT, TTH,THT, HTT, THH, HTH, HHT, HHH} = 8
Favorable outcomes = {TTT, TTH, THT, HTT} = 4

So, the probability of getting at least 2 tails = Favorable outcomes/Total outcomes
= 4/8
= 1/2

Question: ⁷P_m = 210, ⁷C_m = 35, what is the value of m?

Solution:
Given,
⁷P_m = 210
⇒ 7!/(7 - m)! = 210........(1)

⁷C_m = 15
⇒ 7!/m!(7 - m)! = 35..........(2)

(1) ÷ (2) ,
{7!/(7 - m)!}/{7!/m!(7 - m)!} = 210/35
⇒ m! = 6
= 3 × 2 × 1
∴ m = 3

Question: The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?

Solution:
Here,
The order of each letter in the dictionary is ABLORU.

Now,
with A in the beginning, the remaining letters can be permuted = 5! ways.
= 120 ways

Similarly,
with B in the beginning, the remaining letters can be permuted = 5! ways.
= 120 ways

With L in the beginning,
the first word will be LABORU, the second will be LABOUR.

Hence, the rank of the word LABOUR = 5! + 5! + 2
= 120 + 120 + 2
= 242

Question: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is odd?

Solution:
In a simultaneous throw of two dice,
we have n(S) = (6 × 6) = 36

Now,
E= {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5 ,5)}

∴n(E) = 9

∴ P(E) = n(E)/n(S)
= 9/36
= 1/4

Question: In how many ways can a committee of 4 people be formed from 7 men and 6 women such that the number of women is greater than the number of men?

Solution:
Given,
the number of women is greater than the number of men
We have to select women > men

If we want women > men, so there are 2 possible combinations:
1^st combination:
3 women out of 6 women, 1 man out of 7 men = ⁶C₃ × ⁷C₁ = 20 × 7 = 140

2^nd combination:
4 women out of 6 women = ⁶C₄ = 15
                           
∴ Total number of valid committees = 140 + 15 = 155 ways.

Question: How many 8 letter words can be formed by rearranging the letters of the word TRENDING such that T and G occupy the first and last positions respectively?

Solution:
As T and G should occupy the first and last position, the first and last position can be filled in only one following way.
T _ _ _ _ _ _ G.

The remaining 6 positions can be filled in the remaining words (R, E, N, D, I, N) where "N" comes twice.

Total permutations of these 6 letters with one letter repeating = 6!/2! ways

Question: Two cards are drawn at random and without replacement from a standard deck of 52 cards. What is the probability that both cards are face cards?

Solution:
Total card = 52
Total face card = 3 × 4 = 12

Total ways to choose 2 cards from 52 = ⁵²C₂ = (52 × 51)/2 = 1326

Total ways to choose 2 face cards from 12 = ¹²C₂ = (12 × 11)/2 = 66

∴ So, the probability that both cards are face cards = 66/1326
= (2 × 3 × 11)/(2 × 3 × 221)
= 11/221

Question: An art box contains 3 blue pens, 4 green pens, and 5 red pens. In how many ways can a student choose 3 pens such that at least one blue pen is included in the selection?

Solution:
Total number of pens = 3 + 4 + 5
= 12 pens

Total ways to choose any 3 pens from 12 = ¹²C₃
= 220 ways

Ways to choose 3 pens with no blue pen (only green and red pens) = ⁹C₃
= 84 ways 

∴ So, ways to choose 3 pens with at least one blue pen = (220 - 84) ways
= 136 ways

Question: Six friends Rina, Toma, Jui, Lima, Arif, and Sayeed sit randomly in a row of six chairs. What is the probability that Rina and Toma do not sit next to each other?

Solution:
Total number of possibilities = 6! = 720

Number of possibilities where Rina and Toma sit together = 5! × 2! 
= 120 × 2
= 240

So the possibilities where Rina and Toma do not sit together = 720 - 240
= 480

∴Probability that Rina and Toma do not sit next to each other = 480/720
= 2/3

Question: In a meeting, every person shakes hands with every other person exactly once. If the total number of handshakes was 45, how many people were in the meeting?

Solution: 
Let,
the number of people be n.

ATQ,
number of total handshakes,
n(n - 1)/2 = 45
⇒ n(n - 1) = 45 × 2 
⇒ n² - n = 90 
⇒ n² - n - 90 = 0
⇒ n² - 10n + 9n - 90 = 0
⇒ n(n - 10) + 9(n - 10) = 0
⇒ (n - 10)(n + 9) = 0
∴ n = 10, - 9

So the number of people be 10

Question: How many 4 letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', where repetition of letters is not allowed?

Solution:
Here,
'LOGARITHMS' contains 10 different letters.

So the number of word = Number of arrangements of 10 letters, taking 4 at a time
= ¹⁰P₄
= (10 × 9 × 8 × 7)
= 5040

Question: How many 4-digit numbers can be formed from the digits 2, 4, 5, 7, 8, which are divisible by 2 and have no digit repeated?

Solution:
We know,
A number is divisible by 2 if its last digit is even.
The available digits are: 2, 4, 5, 7, 8
The even digits here are: 2, 4, 8
So, the last digit must be one of these 3 digits.
So, Last digit can be chosen in ³C₁ ways 
= 3 ways

As the digit is not repeated
First digit (thousands place) can be chosen in = 4 ways

As the digit is not repeated
Second digit (hundreds place) can be chosen in = 3 ways

As the digit is not repeated
Third digit (tens place) can be chosen in = 2 ways

∴ Total ways = 3 × 4 × 3 × 2 ways
= 72 ways

Question: A volunteer group is organizing a charity event. The group consists of 3 male volunteers and 5 female volunteers. If a team of 4 volunteers is to be chosen at random to lead the event, what is the probability that the team will include at least 2 women?

Solution:
Given,
Total people = 8
∴ Ways of selecting 4 people from 8 = ⁸C₄
= 70

We want at least 2 women, so there are 3 possible combinations:
1^st combination: 2 women and 2 men = ⁵C₂ × ³C₂ = 10 × 3 = 30
2^nd combination: 3 women and 1 man = ⁵C₃ × ³C₁ = 10 × 3 = 30
3^rd combination: 4 women 0 man = ⁵C₄ = 5

∴ Total outcomes = 30 + 30 + 5
= 65

∴ Probability = 65/70
= 13/14

Question: How many distinct arrangements can be made using all the letters of the word "BANANA" such that no two N’s appear together?

Solution:
BANANA has 6 letters, where A = 3 times, N = 2 times and B = 1 time

∴ Total arrangements = 6!/(3! × 2!) = 720/(6 × 2)) = 60

Arrangements where the two N's are together = 5!/3! = (5 × 4 × 3!)/3! = 20

∴ N’s not together = Total arrangements − N’s together
= 60 - 20
= 40​

Question: There are 8 points in a plane, out of which 4 are collinear and the remaining 4 are non-collinear. How many distinct triangles can be formed by joining any 3 of these points?

Solution: 
Total combinations of 3 points from 8 = ⁸C₃
= 56 ways

Given,
there are 4 collinear points

From the 4 collinear points, no triangle can be formed using any 3 of them (since they lie on the same line).
Total combinations of 3 points from the 4 collinear points = ⁴C₃ = 4 ways

So the valid triangles = 56 − 4 = 52​ ways

Question: In an exam, there are 4 multiple choice questions, and each question has 5 choices. Only one answer per question is correct. How many ways can a student fail to get all answers correct?

Solution:
Each question has 5 options, so the total number of ways to answer all 4 questions is = 5⁴
= 5 × 5 × 5 × 5
= 625

Number of ways, getting correct answers = 1⁴ = 1

∴ Number of ways of not getting all answer correct = 625 - 1 = 624

Question: A letter is taken out at random from the word "ENGINEERING", and another is taken out from the word "GREENHOUSE". What is the probability that both selected letters are the same

Solution: 
For E: (3/11) × (3/10) = 9/110
For N: (3/11) × (1/10) = 3/110
For G: (2/11) × (1/10) = 2/110
For R: (1/11) × (1/10) = 1/110

Total probability = (9/110) + (3/110) + (2/110) + (1/110)
= 15/110
= 3/22

Question: Out of 7 consonants and 4 vowels, how many 5-letter words can be formed using 3 consonants and 2 vowels, such that the word always starts with a vowel?

Solution: 
Here,
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (⁷C₃ × ⁴C₂)
= 35 × 6
= 210

Number of groups, each having 3 consonants and 2 vowels = 210

We must arrange them into a 5-letter word starting with a vowel.
From the 2 vowels in each group, choose 1 to be in the first position = 2 ways

Remaining 4 letters can be arranged in 4! = 24 ways

Required number of ways = (210 × 2 × 24) = 10080

Question: In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Solution:
The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in = 3! = 6 ways.

 Required number of ways = (120 × 6) = 720.