Probability, Permutation and Combination

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

Solution:
When three fair (unbiased) coins are tossed, the total number of possible outcomes = 2 × 2 × 2 = 8.
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} 

And all possible outcomes are, {HTT, THT, TTH, TTT} = 4

∴ Probability = (number of favorable outcomes)/(total possible outcomes)
= 4/8
= 1/2

So the probability of getting at least 2 tails is 1/2.

1, 3, 5 এই তিনটি অংক দ্বারা 3! = 6 উপায়ে সংখ্যা গঠন করা যায়
5 দ্বারা বিভাজ্য হতে হলে শেষের অংক 5 রেখে সংখ্যার প্রথম দুটি স্থান বাকি দুইটি অংক দিয়ে 2! = 2 উপায়ে গঠন করা যায়।
∴ সংখ্যাটি 5 দ্বারা বিভাজ্য হবার সম্ভাবনা = 2/6 = 1/3

When at least 2 women are included.
The committee may consist of 3 women, 2 men : It can be done in ⁴C₃ × ⁶C₂ ways
or, 4 women, 1 man : It can be done in ⁴C₄ × ⁶C₁ ways
or, 2 women, 3 men : It can be done in ⁴C₂ × ⁶C₃ ways.
Total number of ways of forming the committees
= ⁴C₂ × ⁶C₃ + ⁴C₃ × ⁶C₂ + ⁴C₄ × ⁶C₁
= 6 x 20 + 4 x 15 + 1 x 6
= 120 + 60 + 6
= 186

Question: In a simultaneous throw of a pair of dice, find the probability of getting a total more than 9.

Solution:
When two fair six-sided dice are thrown together, the total number of possible outcomes = 6 × 6 = 36.
We need the cases where the sum is more than 9, i.e., 10, 11, or 12.
Here are all favorable outcomes, (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6) = 6

∴ Probability = (number of favorable outcomes)/(total possible outcomes)
= 6/36
= 1/6

So the probability of getting a total more than 9 is 1/6.

Question: At what angle the hands of a clock are inclined at 15 minutes past 5?

Solution:
Hours hand moves in 15 past.
5 from 12 p.m = (5 + 15/60) hours = 21/4 hours
Angle of hours hand = (360/12) × (21/4)
= 157.5°

Minutes hands makes angle of = (360/60) × 15
= 90°

Angle between hours and minutes hands = (157.5° - 90°)
= 67.5°

Question: A committee of 4 members is to be formed by selecting out of 7 men and 6 women. In how many different ways can the committee be formed if it should have 3 men and 1 woman?

Solution:
3 men can be selected out of 7 men in
⁷C₃ = 7!/[3!(7 - 3)!]
= (7 × 6 × 5)/(3 × 2 × 1)
= 35 ways

1 woman can be selected out of 6 women in
⁶C₁ = 6 ways

∴ Required number of ways = 35 × 6 = 210

∴ 210 different ways to form the committee with 3 men and 1 woman.

Question: There are 7 non-collinear points. How many triangles can be drawn by joining these points?

Solution: 
A triangle is formed by joining any three non-collinear points in pairs.
There are 7 non-collinear points.

∴ The number of triangles formed =
⁷C₃
= 35

The given word contains 8 different letters.
When the vowels AUE are taken together, we may treat them as 1 letter.
Then,
The letters to be arranged are DGHTR (AUE)
The letters can be arranged in ⁶P₆ = 6!
= 720 ways.
The vowels AUE may be arranged in 3! = 6 ways.
Required number of ways = (720 × 6)
= 4320 ways.

Question: How many words can be formed by using 3 letters from the word 'DELHI'? 

Solution:
Here we will use the Permutations for this question.
We know,
​ ⁿP_r, for this we have,
n = 5, Total 5 Letters
r = 3, Letters word we required

​Now,
ⁿP_r = n!/(n-r)!
= ⁵P₃ 
​= 5!/2! 
​= 120/2 
​= 60

So, Total we can form 60 different permutation of word from Letter Delhi.

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: If ⁶P_r = 120, what is the value of r?

Solution:
আমরা জানি, ⁿP_r = n!/(n - r)!
দেওয়া আছে,
⁶P_r = 120
⇒ 6!/(6 - r)! = 120
⇒ 720/(6 - r)! = 120 (কারণ 6! = 720)
⇒ (6 - r)! = 720/120
⇒ (6 - r)! = 6
⇒ (6 - r)! = 3!   (3! = 3 × 2 × 1 = 6)
⇒ 6 - r = 3
⇒ r = 6 - 3
∴ r = 3

Question: The ratio of the number of red balls to yellow balls to green balls in an urn is 3 : 4 : 5. What is the probability that a ball chosen at random from the urn is a red ball? 

Solution:
লাল, হলুদ, ও সবুজ বলের সংখ্যার আনুপাতিক মান যথাক্রমে 3, 4, 5
মোট বলের সংখ্যার আনুপাতিক মান = 3 + 4 + 5
 = 12

∴ বলটি লাল হওয়ার সম্ভাব্যতা = 3/12
= 1/4

Required no. of numbers = 5 ×⁵P₄
= 5 × 5!
= 5 ×120
= 600

Question: Two unbiased coins are tossed. What is the probability of getting at most one head?

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

∴ Required Probability = 3/4

Question: One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King) only?

Solution:
A standard deck has 52 cards.
Face cards are Jack, Queen, King
There are 4 suits (hearts, diamonds, clubs, spades)
So number of face cards = 3 × 4 = 12

Total possible outcomes (when drawing one card) = 52
Favorable outcomes (drawing a face card) = 12

∴ Probability = favorable outcomes/total outcomes
= 12/52
= 3/13

So the probability that the card drawn is a face card (Jack, Queen, or King) is 3/13. 

Required number of ways,
= ⁶C₄
= 6×5×4! /2!4!
= 15 ways

Question: In how many ways can a committee of 4 men and 3 women be formed from 6 men and 5 women?

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

∴ Number of ways = ⁶C₄ × ⁵C₃
= {6!/4!(6 - 4)!} × {5!/3!(5 - 3)!}
= {(6 × 5)/2} × {(5 × 4)/2}
= 15 × 10
= 150 ways

Question: If ⁿC₁₄ = ⁿC₇, what is the value of ⁿC₂?

Solution:
আমরা জানি,
যদি ⁿC_a = ⁿC_b হয়, তবে হয় a = b অথবা a + b = n হবে।

এখানে,
ⁿC₁₄ = ⁿC₇
⇒ 14 + 7 = n
⇒ n = 21

∴ ⁿC₂ = ²¹C₂
= 21!/(2! × (21 - 2)!) 
= 21!/(2! × 19!) 
= (21 × 20 × 19!)/(2 × 1 × 19!) 
= (21 × 20)/2 
= 21 × 10
= 210

Question: Rakib flipped a fair coin 6 times and got tails every time. What is the probability that he will get a tail on the 7th flip?

Solution:
মুদ্রা নিক্ষেপের প্রতিটি ট্রায়াল বা নিক্ষেপ একটি স্বাধীন ঘটনা (Independent Event)।

রাকিবের আগে ৬ বারই Tails পাওয়া, ৭ম নিক্ষেপের ফলাফলের ওপর কোনো প্রভাব ফেলবে না।

একটি নিরপেক্ষ মুদ্রার (Fair coin) ক্ষেত্রে কেবল দুটি সম্ভাব্য ফলাফল থাকে: Head অথবা Tail।
এখানে মোট ফলাফল সংখ্যা, n(S) = 2

৭ম নিক্ষেপে Tail পাওয়ার অনুকূল ফলাফল সংখ্যা, n(E) = 1

∴ P(E) = n(E)/n(S)
= 1/2

We need to SELECT people.
[SELECT = Combination = ⁿC_r = n!/r!(n-r)!

Here, we first have to select 10 ladies from 19.
Select = Combination
∴ Select 10 ladies = ¹⁹C₁₀

Arrange 10 ladies in circle = 10 - 1 = 9! ways
19 - 10 = 9 ladies remain.
Arrange the remaining 9 ladies in another circle = 9 - 1 = 8! ways

∴ Total ways to arrange 19 ladies in 2 required circles = ¹⁹C₁₀ × 9! × 8!

The letters of the word 'BAKERY' be arranged in 6! ways 
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

Question: In a box, there are 6 white, 4 black, and 2 yellow balls. If two balls are drawn one after the other without replacement, what is the probability that the first one is black and the second one is yellow?

Solution:
মোট বলের সংখ্যা = 6 (সাদা) + 4 (কালো) + 2 (হলুদ) = 12টি

প্রথম বলটি কালো হওয়ার সম্ভাবনা = 4/12 = 1/3

প্রথম বলটি তোলার পর বাক্সে মোট বলের সংখ্যা = 12 - 1 = 11টি

দ্বিতীয় বলটি হলুদ হওয়ার সম্ভাবনা = 2/11

∴ প্রথমটি কালো এবং দ্বিতীয়টি হলুদ হওয়ার সম্ভাবনা:
= (প্রথমটি কালো হওয়ার সম্ভাবনা) × (দ্বিতীয়টি হলুদ হওয়ার সম্ভাবনা)
= (1/3) × (2/11)
= 2/33

Question: In how many ways can 3 guests from a group of 8 guests be seated around a circular table?

Solution:
8 জন থেকে 3 জন নির্বাচন করার উপায়:
⁸C₃ = 8!/(3! × 5!)
= (8 × 7 × 6)/(3 × 2 × 1)
= 336/6
= 56

3 জন ব্যক্তিকে একটি গোলাকার টেবিলে সাজানোর উপায় = (3 - 1)!
= 2!
= 2

∴ মোট উপায় = 56 × 2 = 112

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

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.
Number of 7 digit numbers = 7!3! × 2! = 420
But out of these 420 numbers,
there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.
= 6!3! × 2! = 60
Hence the required number of 7 digits numbers = 420 - 60 = 360

Question: In a city, 40% of the people are illiterate and 60% are poor. Among the rich, 10% are illiterate. What percent of the poor population are illiterate?

Solution:
মনেকরি
মোট জনসংখ্যা = ১০০ জন
অশিক্ষিত লোকের সংখ্যা = ১০০ এর ৪০%৳
= ১০০ এর ৪০/১০০
= ৪০ জন

গরীব লোকের সংখ্যা = ১০০ এর ৬০%
= ১০০ এর ৬০/১০০
= ৬০ জন

ধনী লোকের সংখ্যা = ১০০ এর ৪০%
= ১০০ এর ৪০/১০০
= ৪০ জন

ধনীদের মধ্যে অশিক্ষিত লোকের সংখ্যা = ৪০ এর ১০%
= ৪০ এর ১০/১০০
= ৪ জন

গরীবদের মধ্যে অশিক্ষিত লোকের সংখ্যা = (৪০ - ৪) জন
= ৩৬ জন

৬০ জনের মধ্যে অশিক্ষিত ৩৬ জন
১ জনের মধ্যে অশিক্ষিত ৩৬/৬০ জন
১০০ জনের মধ্যে অশিক্ষিত (৩৬ × ১০০)/৬০ জন
= ৬০ জন

Question: The probability that a card drawn from a pack of 52 cards will be a diamond or a king is -

Solution:
Here, n(S) = 52
There are 13 cards of diamond (including one king) and there are three more kings.
Let E = event of getting a diamond or a king
Then, n(E) = (13 + 3) = 16
∴P(E) = n(E)/(S) = 16/52 = 4/13

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

Solution:
'ARRANGE' শব্দটিতে মোট 7 টি বর্ণ আছে।

এই বর্ণগুলোর মধ্যে
'A' আছে 2 বার,
'R' আছে 2 বার,
এবং বাকি বর্ণগুলো ভিন্ন ভিন্ন।

সুতরাং, মোট সাজানোর সংখ্যা = 7!/(2! × 2!)
= 5040/(2 × 2)
= 5040/4
= 1260

∴ মোট বিন্যাস সংখ্যা = 1260

Question: A box contains 7 red balls, 8 black balls, and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is neither red nor white?

Solution:
Number of red balls = 7
Number of black balls = 8
Number of white balls = 5

∴ Total balls = 7 + 8 + 5 = 20

Let, event E = The ball drawn is neither red nor white, so it must be black.
∴ Number of favorable outcomes = 8

∴ P(E) = 8/20
= 2/5

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

There are a total of 18 balls.
∴ Probability of the balls to be different is,
= (⁴C₁ × ⁶C₁ × ⁸C₁)/(¹⁸C3)
= (4 × 6 × 8)/(18×17×16 / 6)
= (4 × 6 × 8 × 6)/(18 × 17 × 16)
= 4/17

The word PUZZLE has 6 different letters, but ATQ, the vowels should always come together.
Now, let the vowels UE as a single entity.
Therefore, the number of letters is PZZL = 4, and UE = 1
Since the total number of letters = 4+1 = 5

So the arrangement would be in ⁵P₅ = 5!/(5 - 5)!
= 5!/0!
= 5!
= 5 × 4 × 3 × 2 × 1
= 120 ways

Now, the vowels AE can arrange in 2 different ways, i.e., ²P₂ = 2! = 2 × 1 = 2 ways
Hence, the new words that can be formed = 120 × 2 = 240.

Question: Three dice are thrown together. Find the probability of getting a total of at least 6?

Solution:
The total possible ways of sum of 3 dice less than, 6
(1 + 1 + 1), (1 + 1 + 2), (1 + 1 + 3), (1 + 2 + 1), (1 + 2 + 2), (1 + 3 + 1), (2 + 1 + 1), (2 + 1 + 2), (2 + 2 + 1), (3 + 1 + 1)

Total 10 ways.
Total Sample space = 6³ = 216

Probability of getting the sum less than 6 = 10/216
Probability of getting the sum at least 6 = 1 - (10/216)
= (216 - 10)/216
= 206/216
= 103/108

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 × 4C2)
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5! = 120
Required number of ways = (210 x 120)
= 25200.

Given, Green marbles = 2
Red marbles = 2
Total marbles = 2 + 2 = 4
Probability of randomly picked marbles that both of them are green = ²C₂ / ⁴C₂ = 1/6

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

We want 2 orange towels
That means to choose one AND then choose other from the remaining towels
There are 6 orange towels
Total 15 towels

So probability for two orange towels = 6/15 × 5/14
= 1/7 [Here we reduce the denominator i.e. the total number of towels because once we remove a towel from the box we do not put it back in the box.
So while removing the 2^nd towel, there are only 15 - 1 = 14 towels in the box.]