IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

Question: Two dice are thrown simultaneously. What is the probability that the total score is an even number?

Solution:
Clearly, n(S) = (6 × 6) = 36

Let E = Event that the sum is an even number.
E = {(1, 1), (1, 3), (1, 5),
     (2, 2), (2, 4), (2, 6),
     (3, 1), (3, 3), (3, 5),
     (4, 2), (4, 4), (4, 6),
     (5, 1), (5, 3), (5, 5),
     (6, 2), (6, 4), (6, 6)}

∴ n(E) = 18

∴ P(E) = n(E)/n(S)
= 18/36
= 1/2

Question: A card is drawn from a pack of 52 cards. The probability of getting a queen of spades or a king of diamonds is:

Solution:
Here, n(S) = 52
Let E = event of getting a queen of spades or a king of diamonds.
Then, n(E) = 2

∴ P(E) = n(E)/n(S)
= 2/52
= 1/26

Question: ⁶P_m = 120, ⁶C_m = 20, what is the value of m?

Solution:
Given,
⁶P_m = 120
⇒ 6!/(6 - m)! = 120 ..........(1)

⁶C_m = 20
⇒ 6!/{m!(6 - m)!} = 20 ..........(2)

(1) ÷ (2),
{6!/(6 - m)!} / [{6!/{m!(6 - m)!}] = 120/20
⇒ m! = 6

We know,
3! = 3 × 2 × 1 = 6
∴ m = 3

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

Solution:
The word 'ORATION' contains 7 letters, where O is repeated twice.

The vowels are O, A, I, O.

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

Then, we have to arrange the letters RTN (OAIO).

Now, 4 letters can be arranged in 4! = 24 ways.

The vowels (O, A, I, O) can be arranged among themselves in = 4!/2! = 12 ways.

∴ Required number of ways = (24 × 12) = 288.

Question: In an exam, there are 3 multiple choice questions, and each question has 4 choices. Only one answer per question is correct. How many ways can a student fail to get all answers correct?

Solution:
Each question has 4 options, so the total number of ways to answer all 3 questions is = 4³
= 4 × 4 × 4
= 64

Number of ways, getting correct answers = 1³ = 1

∴ Number of ways of not getting all answers correct = 64 - 1 = 63

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

Solution:
MAMMAL has 6 letters, where M = 3 times, A = 2 times and L = 1 time.
First arrange the letters other than M.
Number of arrangements of A, A, L = 3!/2! = 3

Now place the three M's in the gaps of these letters.
For example: _ A _ A _ L _
Total gaps = 4

Number of ways to choose 3 gaps for M = ⁴C₃ = 4

∴ Required number of arrangements = 3 × 4
= 12

Question: A bag contains 5 red balls, 7 blue balls and 6 green balls. One ball is drawn at random and replaced with 2 green balls. What is the probability that the first ball drawn was either red or blue and the second drawn was green in colour?

Solution:
Number of Red balls = 5  
Number of Blue balls = 7  
Number of Green balls = 6  

Total number of balls = 5 + 7 + 6 = 18  

After the first ball is drawn, it is replaced with 2 green balls.
So the total number of balls becomes 18 - 1 + 2 = 19,
and the number of green balls becomes 6 + 2 = 8.

So, Required probability  
= (5/18) × (8/19) + (7/18) × (8/19)
= 8/19 [ (5/18) + (7/18)]
= (8/19) × (12/18)
= 16/57

Question: In how many ways can 4 students be chosen from a class of 12 students?

Solution:
Here, total number of students, n = 12
Number of students to be chosen, r = 4
We know, the number of ways to choose r objects from n objects is ⁿC_r
So, total ways = ¹²C₄
= 12!/{4! × (12 - 4)!}
= 12!/(4! × 8!)
= (12 × 11 × 10 × 9 × 8!)/(4 × 3 × 2 × 1 × 8!)
= (12 × 11 × 10 × 9)/24
= 11880/24
= 495

Question: A volunteer group consists of 4 male volunteers and 6 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 = 10

∴ Ways of selecting 4 people from 10 = ¹⁰C₄
= 210

We want at least 2 women, so there are 3 possible combinations:
1st combination: 2 women and 2 men
= ⁶C₂ × ⁴C₂
= 15 × 6
= 90

2nd combination: 3 women and 1 man
= ⁶C₃ × ⁴C₁
= 20 × 4
= 80

3rd combination: 4 women and 0 man
= ⁶C₄
= 15

∴ Total outcomes = 90 + 80 + 15
= 185

∴ Probability = 185/210
= 37/42

Question: How many 4-digit numbers can be formed from the digits 1, 3, 4, 6, 9, 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: 1, 3, 4, 6, 9
Even digits here are: 4, 6
So, the last digit must be one of these 2 digits.

So, last digit can be chosen in 2 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 = 2 × 4 × 3 × 2 ways
= 48 ways

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

Solution:
Here,
'TRIANGLE' contains 8 different letters.

So the number of words = Number of arrangements of 8 letters, taking 4 at a time
= ⁸P₄
= (8 × 7 × 6 × 5)
= 1680

Question: Two unbiased coins are tossed. What is the probability of getting at least 1 tail?

Solution:
Total outcomes = {TT, TH, HT, HH} = 4
Favorable outcomes = {TT, TH, HT} = 3

So, the probability of getting at least 1 tail = Favorable outcomes/Total outcomes
= 3/4

Question: How many different registration numbers can be formed using two distinct letters followed by two distinct digits?

Solution:
Here,
Number of ways to choose and arrange two distinct letters out of 26 alphabets = ²⁶P₂  
= 26 × 25  
= 650

Number of ways to choose and arrange two distinct digits out of 10 digits (0 - 9) = ¹⁰P₂  
= 10 × 9  
= 90

Total number of registration numbers = 650 × 90  
= 58,500

Question: In how many ways can the letters of the word 'MISSISSIPPI' be arranged such that the first letter is always 'M'?

Solution:
The word 'MISSISSIPPI' contains 11 letters.

Condition: The first letter is fixed as 'M'.

So, the remaining 11 - 1 = 10 positions are to be filled by the remaining letters.

Among the remaining 10 letters, the repeated letters are:
I (4 times), S (4 times), P (2 times).

∴ Number of arrangements of the remaining 10 letters
= 10!/(4! × 4! × 2!)
= 3,628,800/(24 × 24 × 2)
= 3,628,800/1,152
= 3,150

Question: In how many ways can 3 guests from a group of 6 guests be seated around a circular table?

Solution:
Ways of selecting 3 guests from 6 guests:
⁶C₃ = 6!/(3! × 3!)
= (6 × 5 × 4)/(3 × 2 × 1)
= 120/6
= 20

Ways of arranging 3 persons around a circular table = (3 - 1)!
= 2!
= 2

∴ Total ways = 20 × 2 = 40

Question: All distinct permutations of the letters of the word PRIDE are arranged in alphabetical order. What is the rank of the word PRIDE?

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

Now,
with D in the beginning, the remaining letters can be permuted = 4! ways.
= 24 ways

Similarly,
with E in the beginning, the remaining letters can be permuted = 4! ways.
= 24 ways

Similarly,
with I in the beginning, the remaining letters can be permuted = 4! ways.
= 24 ways

Now,
with P in the beginning and R in the second position,
the remaining letters are D, E, I.

Before R, there are D, E, I.

So permutations = 3 × 3!
= 3 × 6
= 18

Now,
with P and R fixed, the remaining letters are D, E, I.

Before I, there are D, E.

So permutations = 2 × 2!
= 2 × 2
= 4

Finally,
add the word itself.

Hence, the rank of the word PRIDE
= 24 + 24 + 24 + 18 + 4 + 1
= 95

Question: In a meeting, every person shakes hands with every other person exactly once. If the total number of handshakes was 28, how many people were in the meeting?

Solution:
Let,
the number of people be n.

ATQ,
number of total handshakes,
n(n - 1)/2 = 28
⇒ n(n - 1) = 28 × 2
⇒ n² - n = 56
⇒ n² - n - 56 = 0
⇒ n² - 8n + 7n - 56 = 0
⇒ n(n - 8) + 7(n - 8) = 0
⇒ (n - 8)(n + 7) = 0

∴ n = 8, - 7

So, the number of people be 8.

Question: A two member committee comprising of one male and one female member is to be constituted out of six males and four females. Amongst the females, Ms. C refuses to be a member of the committee in which Mr. D is taken as the member. In how many different ways can the committee be constituted?

Solution:
Total number of committees without restriction
= ⁶C₁ × ⁴C₁
= 6 × 4
= 24

Now,
the committee containing Mr. D and Ms. C is not allowed.
Number of such committees = 1

∴ Required number of committees
= 24 - 1
= 23

Question: A box contains 24 electric bulbs, out of which 6 are defective. Two bulbs are chosen at random from this box. The probability that at least one of them is defective is:

Solution:
Given that,
Total bulbs = 24
Defective bulbs = 6
Non-defective bulbs = 24 - 6 = 18
Two bulbs are chosen at random (without replacement)

Now,
P(both non-defective) = (18/24) × (17/23)
= 306/552
= 51/92

And,
∴ P(at least one defective)
= 1 - P(both non-defective)
= 1 - (51/92)
= (92 - 51)/92
= 41/92

Question: In how many ways can 8 examination papers be arranged so that the best and the worst papers never come together?

Solution:
No. of ways in which 8 papers can be arranged = 8! ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 7 papers.
These 7 papers can be arranged in 7! ways.
And two papers can be arranged themselves in 2! ways.

No. of arrangements when best and worst papers do not come together
= 8! - 7! × 2!
= 7!(8 - 2)
= 6 × 7!
= 30,240