Probability, Permutation and Combination

n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
n(E)= 10 + 8 - 1 (since 2 is common in both item)
= 17
P(E U P) = n(E)/n(s)
= 17/20 + 8/20 - 1/20
= 17/20

Question: A box contains 5 green, 3 yellow, and 2 black balls. If one ball is drawn at random, what is the probability that it will not be yellow?

Solution:
Green balls = 5
Yellow balls = 3
Black balls = 2

∴ Total balls = 5 + 3 + 2 = 10
Number of non-yellow balls = Green + Black = 5 + 2 = 7

We know,
Probability,
P(not yellow) = favorable outcomes/total outcomes
= 7/10

∴ The probability that the drawn ball will not be yellow is 7/10.

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

Solution:
'ENGINE' শব্দটিতে মোট বর্ণ আছে 6টি।

এখানে,
E আছে ২টি
N আছে ২টি

আমরা জানি, পুনরাবৃত্ত বর্ণ থাকলে বিন্যাস সংখ্যা = n!/(p! × q!)
∴ বিন্যাস সংখ্যা = 6!/(2! × 2!)
= (6 × 5 × 4 × 3 × 2 × 1)/{(2 × 1) × (2 × 1)}
= 720/(2 × 2)
= 720/4
= 180

Question: In a box, there are 4 red, 5 blue and 7 green balls. One ball is picked up randomly. What is the probability that it is neither red nor blue?

Solution: 
 
মোট বলের সংখ্যা, n(S) = 4 + 5 + 7
= 16

ধরি, লাল বা নীল না হওয়ার সম্ভাবনা = P(E)
সবুজ হওয়ার সম্ভাবনা = n(G) = 7

∴ লাল বা নীল না হওয়ার সম্ভাবনা = P(E) 
= n(G)/n(S)
= 7/16

Question: In a class, there are 10 boys and 8 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is:

Solution: 
Total students = 10 + 8 = 18
Let S be the sample space, and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number of ways of selecting 3 students out of 18 = ¹⁸C₃
= (18 × 17 × 16)/(3 × 2 × 1)
= 816

And, n(E) = Number of events of selecting 1 girl and 2 boys = ⁸C₁ × ¹⁰C₂ 
= 8 × [(10 × 9)/2]
= 8 × 45
= 360

∴ Probability = n(E)/n(S)
= 360/816
= 15/34

A team of 6 members has to be selected from the 10 players.
This can be done in 10C6 or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore,
total ways the selection can be made is = 210 × 6
= 1260

Question: How many ways can the letters of the word 'LETTER' be arranged so that the two T’s are together?

Solution:
The word LETTER has 6 letters.
L, E, T, T, E, R
Here, 2 T’s are identical, 2 E’s are identical, and L and R appear once each.

Now,
Treat the two T’s as a single unit (bundle them together).
So we now have 5 units to arrange: (TT), L, E, E, R

Total units = 5 and E is repeated 2 times
∴ Number of distinct arrangements = 5!/2!
= 120/2
= 60
Inside the (TT) bundle, the two T’s are identical, so there is only 1 way to arrange them inside the bundle (TT = TT).

Therefore, total number of arrangements where the two T’s are together 60.

We know,
Probability = what we want/Total
Or = add; AND = multiply

We want black OR red pebble
There are 6 black and 8 red pebbles
Total pebbles = 2 + 6 + 4 + 8 = 20

So, Probability = 6/20 + 8/20 = 14/20 = 7/10.

Question: In how many ways can a group of 3 teachers and 4 students be formed from 6 teachers and 9 students?

Solution:
We have 6 teachers and 9 students.
We need to choose 3 teachers from 6 and 4 students from 9.

∴ Number of ways = ⁶C₃ × ⁹C₄
= {6!/(3!(6 - 3)!)} × {9!/(4!(9 - 4)!)}
= {6!/(3! × 3!)} × {9!/(4! × 5!)}
= {(6 × 5 × 4)/(3 × 2 × 1)} × {(9 × 8 × 7 × 6)/(4 × 3 × 2 × 1)}
= 20 × 126
= 2520 ways

Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10
Probability of getting prize P(E) = n(E)/n(S)
= 10/35
= 2/7
Probability of not getting prize = 1 - (2/7)
= 5/7

Total number of events occure when the dice are thrown = 6 × 6 × 6 = 216
Let A be the event of getting a total of at least 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.

So, B = {(1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1)}
Favorable number of cases = 10
So, P(B) = 10/216 = 5/108
We know, P(A) + P(B) = 1
   ∴ P(A) = 1 – P(B)
             = 1 – (5/108)
             = 103/108

Question: If 12 people attend a meeting and each person shakes hands with every other person exactly once, find the total number of handshakes.

Solution:

Total handshakes = ¹²C₂ = 12!/2!(12 - 2)!
= (12 × 11 × 10!)/(2 × 10!)
= 66

∴ The total number of handshakes is 66.

Question: In how many ways can the letters of 'PARALLEL' be arranged if the first letter is always P?

Solution:
The word 'PARALLEL' has a total of 8 letters.

Condition, The first letter is fixed as 'P'.
Now, the remaining 8 - 1 = 7 positions need to be filled with the remaining letters.

Among these 7 letters, there are repeated letters:
A (2 times) and L (3 times)

∴ Number of arrangements for the remaining 7 letters= 7!/(3! × 2!)
= 5040/12
= 420

Question: How many words can be formed by using the letters from the word 'DRIVER' such that all the vowels are never together?

Solution:
We assume all the vowels to be a single character, i.e., 'IE' is a single character.
So, now we have 5 characters in the word, namely, D, R, V, R, and IE.

But, R occurs 2 times.
Number of possible arrangements = 5!/2! = 60

Now, 
​the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together = 60 × 2 = 120

Total number of possible words = 6!/2! = 720/2 = 360

Therefore, the total number of possible words such that the vowels are never together = 360 - 120 = 240

Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect.
Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.
Therefore, the total number of parallelograms formed = ⁷C₂ x ⁶C₂ = 315

Question: Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. Then the total number of persons in the room is-

Solution: 
⇒n(n - 1)/2 = 66
⇒n(n - 1) = 66 × 2 
n² - n = 132 
n² - n - 132 = 0
n² - 12n + 11n - 132 = 0
n(n - 12) + 11(n - 12) = 0
(n - 12)(n + 11) = 0
n = 12, - 11 

Question: There are 8 multiple-choice questions in an examination, and each question has 4 options. In how many ways can these questions be answered?

Solution:
একটি বহু নির্বাচনী প্রশ্নে উত্তর দেওয়ার সম্ভাব্য উপায় সংখ্যা হলো 4টি (কারণ প্রতিটি প্রশ্নে 4টি করে বিকল্প আছে)।

যেহেতু প্রশ্ন সংখ্যা 8টি এবং প্রতিটি প্রশ্নের উত্তর একে অপরের উপর নির্ভরশীল নয়, তাই মোট সম্ভাব্য উপায় সংখ্যা হবে প্রতিটি প্রশ্নের উত্তরের উপায়গুলোর গুণফল।

প্রথম প্রশ্নের সম্ভাব্য উপায় = 4
দ্বিতীয় প্রশ্নের সম্ভাব্য উপায় = 4
............
অষ্টম প্রশ্নের সম্ভাব্য উপায় = 4

∴ নির্ণেয় মোট উপায় = 4 × 4 × 4 × … (8 বার) = 4⁸
অতএব, 8টি বহু নির্বাচনী প্রশ্নের উত্তর 4⁸ উপায়ে দেওয়া যেতে পারে।

Total number of possibilities = 36
Event of getting a doublet = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6
Thus, probability of getting a doublet = 6/36 = 1/6

Question: In how many ways can a group of 3 teachers and 4 students be formed from 6 teachers and 10 students?

Solution:
We have 6 teachers and 10 students.
We need to choose 3 teachers from 6 and 4 students from 10.

∴ Number of ways = ⁶C₃ × ¹⁰C₄
= 6!/{3!(6 - 3)!)} × 10!/{4!(10 - 4)!)}
= 6!/(3! × 3!) × 10!/(4! × 6!)
= {(6 × 5 × 4)/(3 × 2 × 1)} × {(10 × 9 × 8 × 7)/(4 × 3 × 2 × 1)}
= 20 × 210
= 4200 ways

Question: Out of 6 persons working on a project, 2 are graduated. If 2 are selected, what is the probability that there is at least one graduate among them?

Solution:
এখানে মোট ব্যক্তি = 6 জন, গ্র্যাজুয়েট = 2 জন এবং নন-গ্র্যাজুয়েট = (6 - 2) = 4 জন।

6 জন থেকে 2 জনকে বাছাই করার মোট উপায় = ⁶C₂
= (6 × 5)/(2 × 1)
= 15 

এখন, 2 জনের মধ্যে একজনও গ্র্যাজুয়েট না থাকার (অর্থাৎ 2 জনই নন-গ্র্যাজুয়েট হওয়ার) উপায় = ⁴C₂
= (4 × 3)/(2 × 1) = 6 

∴ একজনও গ্র্যাজুয়েট না থাকার সম্ভাবনা = 6/15 = 2/5

আমরা জানি, কমপক্ষে একজন গ্র্যাজুয়েট থাকার সম্ভাবনা = 1 - (একজনও গ্র্যাজুয়েট না থাকার সম্ভাবনা)
= 1 - (2/5)
= 3/5

Question: In how many ways can a team of 4 people be selected from 8 people?

Solution:
Total number of people, n = 8
Number of team members, r = 4

The number of ways to choose the team = ⁿC_r =
⁸C₄ = 8!/[4! × (8 - 4)!]
= (8 × 7 × 6 × 5 × 4!)/(4 × 3 × 2 × 1 × 4!)
= 70

Question: There are 5 red, 4 white, and 3 blue balls in a box. If 3 balls are drawn at random from the box, what is the probability that all three are red?

Solution:
Total number of balls = 5 + 4 + 3 = 12 balls
We are drawing 3 balls at random without replacement.
Total number of ways to choose 3 balls from 12
= ¹²C₃ 
= 12!/(3! × 9!)
= (12 × 11 × 10)/(3 × 2 × 1)
= 220

And
Number of favorable ways (all 3 balls are red)
= Number of ways to choose 3 red balls from 5 red balls
= ⁵C₃
= 5!/(3! × 2!)
= (5 × 4 × 3)/(3 × 2 × 1)
= 10

∴ Probability = (favorable outcomes)/(total outcomes)
= 10/220
= 1/22
So the probability that all three balls are red is 1/22.

Alternatively, using sequential probability (without replacement):
P(1st Red) = 5/12
P(2nd Red) = 4/11 (4 red left out of 11 total)
P(3rd Red) = 3/10 (3 red left out of 10 total)

∴ Total Probability = (5/12) × (4/11) × (3/10) 
= 60/1320
= 1/22. 

Question: What is the probability of getting a sum of 9 when a die is thrown twice?

Solution: 
In two throws of a dice, n(S) = (6 x 6) = 36
Let E = the event of getting a sum
= {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(E) = 4

Hence, P(E) = n(E)/n(S)
= 4/36
= 1/9

Question: There are 10 true-false questions in an examination, these questions can be answered in-
(পরীক্ষায় ১০টি সত্য-মিথ্যা প্রশ্ন আছে, এই প্রশ্নগুলোর উত্তর দেওয়া যেতে পারে কত উপায়ে?)

Solution: 
Total number of question = 10 
Each question has 2 answer.

∴ These question can be answered in 2¹⁰ ways = 1024 ways

Question: A team of 4 men and 3 women is to be formed from 6 men and 5 women. In how many ways can the team be formed?

Solution:
We have 6 men and 5 women.
We need to choose 4 men from 6 and 3 women from 5.

Number of ways to choose 4 men from 6:
⁶C₄ = 6!/(4!(6 - 4)!) 
= (6 × 5)/(2 × 1)
= 15

Number of ways to choose 3 women from 5:
⁵C₃ = 5!/(3!(5 - 3)!)
= (5 × 4)/(2 × 1)
= 10

Total number of ways to form the team = 15 × 10 = 150

Therefore, the team can be formed in 150 different ways.

Question: A team of 6 players is to be selected from 8 batsmen and 5 bowlers. How many ways can this be done if exactly 4 batsmen must be selected?

Solution:
এখানে, একটি দল গঠন করতে হলে,  4 জন ব্যাটসম্যান এবং 2 জন বোলার নির্বাচন করতে হবে।

8 জন ব্যাটসম্যান থেকে 4 জন ব্যাটসম্যান নির্বাচন করার উপায়:
⁸C₄ = 8!/{4! (8 - 4)!} 
= (8 × 7 × 6 × 5)/(4 × 3 × 2 × 1)
= 70 টি

5 জন বোলার থেকে 2 জন বোলার নির্বাচন করার উপায়:
⁵C₂ = 5!/{2! (5 - 2)!}
= (5 × 4)/(2 × 1)
= 10 টি

সুতরাং, মোট সম্ভাব্য উপায় = 70 × 10 = 700 টি।

অতএব, দল গঠনের মোট উপায় হলো 700 টি।

The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6)
= 720.

প্রথম স্থানে 9 স্থির রেখে তৃতীয় স্থানে 3 নির্দিষ্ট করে সাজানো যায় ⁸P₃ উপায়ে
প্রথম স্থানে 9 স্থির রেখে তৃতীয় স্থানে 6 নির্দিষ্ট করে সাজানো যায় ⁸P₃ উপায়ে
∴ মোট বিন্যাস সংখ্যা =  ⁸P₃ +  ⁸P₃ 
^{= (8×7×6×5! / 5!) + (8×7×6×5! / 5!)
= 336 + 336
= 672}

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.

ⁿC_r= n!/(r!) (n – r)!
3 consonants out of 5 can be selected in ⁵C₃ ways
2 vowels out of 4 can be selected in ⁴C₂ ways
In ⁵C₃ and ⁴C₂ ways the consonants and vowels can be selected.
Remember ⁵C₃ and ⁴C₂, hence multiply the terms ⁵C₃ × ⁴C₂
⁵C₃ × ⁴C₂ = {5!/3!(5 - 3)!} × {4!/2!(4 - 2)!}
= {(5 × 4 × 3 × 2 × 1)/(3 × 2 × 2)} × (4 × 3 × 2 × 1)/(2 × 2)
=(5 × 2) × (3 × 2)
60 groups can be made each having 3 consonants and 2 vowels
Each word contains 3 consonants and 2 vowels i.e 5 letters.

If N different objects are to be arranged, then they can be arranged in N! ways.
5 letters can be arranged in 5! Ways.
5! = 5 x 4 x 3 x 2 x 1 = 120
Required number of words = (120 x 60) = 7200

Question: Two dice are tossed at once. What is the probability that their combined total is no less than 11?

Solution:
If two dice are rolled together total possible outcomes = 6 × 6 = 36

But we want sum ≥ 11

Sum of the pairs of dice , 
Sum =  11 where (5, 6), (6, 5) here 2 outcomes
Sum = 12 (6, 6) here 1 outcomes

∴ Total outcomes = 2 + 1 = 3

The probability = 3/36
= 1/12

Question: How many permutations of the letters of the word BALLOON are there?

Solution:
BALLOON শব্দটিতে মোট অক্ষর সংখ্যা = 7 টি
যার মধ্যে,
L আছে 2 বার
O আছে 2 বার

∴ Permutations = 7!/(2! × 2!)
= (7 × 6 × 5 × 4 × 3 × 2)/(2 × 2)
= 7 × 6 × 5 × 2 × 3 
= 1260

Question: Out of 60 books in a school library, 18 belong to the science fiction genre. If a student chooses one book randomly, find the probability it isn’t a science fiction book.

Solution:
Probability of picking a book that is a science fiction book = 18/60
= 3/10

Probability of picking a book that is not a science fiction book = (1 - 3/10)
= (10 - 3)/10
= 7/10

When the owner of the ball to be included always, we have to select 10 players out of 14.

The required no. of ways
¹⁴C₁₀ = 14!/(10!4!) 
= 14.13.12.11 / 4.3.2.1
= 7.13.11
= 1001

টসে প্রথমবারে হেড আসার সম্ভাবনা 1/2
এবং পরেরবারে টেল আসার সম্ভাবনা 1/2
সুতরাং, প্রথম টসে হেড এবং পরের টসে টেল আসার সম্ভাবনা = 1/2 × 1/2 = 1/4

Question: A bag contains 8 red marbles, 12 yellow marbles, and 4 purple marbles. What is the probability of drawing a yellow marble?

Solution:
Total number of marbles = 8 + 12 + 4
= 24

And number of yellow marbles (favorable outcomes) = 12

Probability of drawing a yellow marble = (Number of yellow marbles)/(Total number of marbles)
= 12/24
= 1/2

So the probability is 1/2.

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: Two dice are thrown simultaneously. What is the probability of getting the face numbers are same?

Solution:
In a simultaneous throw of two dice, we have n(S) = 6 × 6 = 36
Let E = event of getting two numbers are same.
Then E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
therefore, n(E) = 6
And P(E) = P (getting two numbers are same)

∴ P(E) = n(E)/n(S)
= 6/36
= 1/6

Hence the answer is 1/6 .

Question: If the chairperson's seat is fixed, in how many ways can 6 people be seated at a circular table?

Solution:
Chairperson’s seat is fixed
∴ Remaining 5 people can be arranged in 5! ways
= 5 × 4 × 3 × 2 
= 120 ways

There are 12 celebrities. A handshake needs 2 people.
This simply means in how many ways 2 people can be selected out of 12.

So the answer is ¹²C₂

ⁿC_r = n!/r!(n - r)!

∴ ¹²C₂ = 12!/2!(12 - 2)! = (12 × 11)/2
= 66 = number of handshakes. [If there are n people and they shake hands only once with each other, then, Number of handshakes = ⁿc₂ = n(n -1)/2]

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