Probability, Permutation and Combination

Total number of kings is 4 out of 52 cards.
Total number of queens is 4 out of 52 cards
Number of favorable outcomes i.e. ‘a king or a queen’ is 4 + 4 = 8 out of 52 cards.
Therefore, probability of getting ‘a king or a queen’ = Number of favorable outcomes
∴ P(E) = Total number of possible outcomes = 8/52 = 2/13

Question: In how many ways can a leap year have 53 Mondays?

Solution:
In a leap year there are 366 days
 52 weeks + 2 extra days.
So to have 53 Mondays one of these two days must be a Monday.
This can occur in only 2 ways.
 (Sunday and Monday) or (Monday and Tuesday).
Thus number of ways = 2

Question: Three coins are tossed. Find the probability of exactly 2 heads -

Solution: 
n(S) = 2³ = 8 
Let E = Event of getting exactly two heads,
= {(H, H, T), (H, T, H), (T, H, H)}
= n(E)
= 3

Required probability = 3/8

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

Solution:
মোট বলের সংখ্যা = 4 (লাল) + 6 (সাদা) + 2 (কালো) = 12টি।
∴ প্রথম বলটি লাল হওয়ার সম্ভাবনা = 4/12 = 1/3

যেহেতু বলটি প্রতিস্থাপন করা হয়নি (Without replacement),
তাই প্রথম বলটি তোলার পর বক্সে মোট বলের সংখ্যা = 12 - 1 = 11টি।
∴ দ্বিতীয় বলটি সাদা হওয়ার সম্ভাবনা = 6/11

∴ প্রথমটি লাল এবং দ্বিতীয়টি সাদা হওয়ার সম্ভাবনা = (প্রথমটি লাল হওয়ার সম্ভাবনা) × (দ্বিতীয়টি সাদা হওয়ার সম্ভাবনা)
= (1/3) × (6/11)
= 6/33
= 2/11

∴ নির্ণেয় সম্ভাবনা 2/11

Question: All possible three-digit numbers are formed by the digits 1, 2, 3, 5, 6 (without repetition). If one number is chosen randomly, what is the probability that it is divisible by 5?

Solution:
প্রদত্ত অঙ্কগুলো: 1, 2, 3, 5, 6

তাহলে, তিন অঙ্কের মোট সংখ্যা = ⁵P₃ = 5 × 4 × 3 = 60 টি।

একটি সংখ্যা 5 দ্বারা বিভাজ্য হবে যদি তার শেষ অংকটি 5 হয়।
• শেষ অংকটি 5 (১টি উপায়)।
• বাকি চারটি অংক (1, 2, 3, 6) থেকে প্রথম দুটো স্থান পূরণ করতে হবে।
• প্রথম দুটো স্থান পূরণ করার উপায় = ⁴P₂ = 4 × 3 = 12 টি।

সুতরাং, 5 দ্বারা বিভাজ্য মোট সংখ্যা = 12 টি।

অতএব, সংখ্যাটি 5 দ্বারা বিভাজ্য হবার সম্ভাবনা = (অনুকূল ফলাফল)/(মোট ফলাফল)
= 12/60
= 1/5

Question: In how many ways can a group of 4 men and 3 women be made out of a total of 6 men and 5 women?

Solution:
There are 6 men and 5 women. We have to select 4 men out of 6 and 3 women out of 5.

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

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

∴ The number of ways of making the selection = 15 × 10 = 150 ways

Total number of balls, n(S) = 8 + 7 + 6 = 21
n(E) = Number of ways in which a ball can be selected which is neither yellow nor black = 7 (∵ there are only 7 balls which are neither yellow nor black)
P(E) = n(E)/n(S) = 7/21 = 1/3

Question: How many triangles can be formed with 18 points?

Solution:
আমরা জানি,
একটি ত্রিভুজ গঠন করতে ৩ টি বিন্দু প্রয়োজন হয়।
তাহলে,
18 টি বিন্দু দিয়ে গঠিত ত্রিভুজের সংখ্যা = ¹⁸C₃
= 18!/{3! × (18 - 3)!}
= 18!/(3! × 15!)
= (18 × 17 × 16 × 15!)/(3 × 2 × 1 × 15!)
= (18 × 17 × 16)/(3 × 2 × 1)
= (18 × 17 × 16)/6
= 4896/6
= 816

∴ 18 টি বিন্দু দিয়ে মোট 816 টি ত্রিভুজ গঠন করা যাবে।

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!   (কারণ 3! = 3 × 2 × 1 = 6)
⇒ 6 - r = 3
⇒ r = 6 - 3
∴ r = 3

Question: In how many different ways can the letters of the word ‘BALLOON’ be arranged?

Solution:
Number of letters in the word = 7

Repeated letters:
L = 2 times
O = 2 times
The remaining letters (B, A, N) are different.

∴ Number of arrangements
= 7!/(2! × 2!)
= 5040/4
= 1260

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

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

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

Here S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting the multiple of 3
Then,
E = {3,6}
P(E) = n(E)/n(S)
= 2/6
= 1/3

No. of boundaries = 6
No. of balls = 30
No. of balls without boundaries = 30 – 6 =24
Probability of no boundary = 24/30 = 4/5

Question: In how many ways can 5 boys and 4 girls be seated in a row so that they sit alternately?

Solution:
Let the arrangement be,
B G B G B G B G B

5 boys can be seated in 5! ways
4 girls can be seated in 4! ways

∴ Required number of ways
= 5! × 4!
= 120 × 24
= 2880

Question: In a box, there are 5 red, 9 blue, and 10 green balls. One ball is picked randomly. What is the probability that it is neither red nor green?

Solution:
Total number of balls = 5 + 9 + 10 = 24

Let, E be the event that the ball is neither red nor green = the event that the ball is blue.
∴ n(E) = 9

∴ Probability = Number of favorable outcomes/Total outcomes
= 9/24
= 3/8

Question: There are 35 people in a room. If each person shakes hands with every other person exactly once, how many handshakes will take place in total?

Solution:
Total handshakes = ³⁵C₂
= (35 × 34)/2
= 595

∴ Total handshakes = 595

৮ জনের মধ্যে ৪ জন মহিলা ও ৪ জন পুরুষ।
দুইজনকে দৈবভাবে নিলে মহিলা আসার সম্ভাবনা বের করতে হবে।
প্রথম ১ জন নিলে মহিলা আসার সম্ভাবনা = মোট মহিলা/মোট সংখ্যা
= ৪/৮
= ১/২
তখন মোট সংখ্যা থাকবে ৭ জন ও মহিলা থাকবে ৩ জন।
তারপর ১ জন নিলে মহিলা আসার সম্ভাবনা = ৩/৭।
তাহলে দুইজন নিলে একত্রে সম্ভাবনা হবে = ১/২ × ৩/৭
= ৩/১৪।

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

One person can select one house out of 3 = ³C₁ ways =3 ways
Hence, three persons can select one  house in = 3 × 3 × 3 = 27 ways
So, All three apply for the same house, its probability = 3/27 = 1/9

Question: A college has 10 basketball players. A 5 member's team and a captain will be selected out of these 10 players. How many different selections can be made?

Solution:
We can select the 5 member team out of the 10 in =  ¹⁰C₅ ways
= 252 ways

The captain can be selected from amongst the remaining 5 players in 5 ways.

∴ The total ways the selection of 5 players and a captain can be made = 252 × 5 ways
= 1260 ways

Question: Numbers from 10 to 20 are written on cards. One card is drawn at random. What is the probability that the number is prime or divisible by 3?

Solution:
The numbers are, 
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
∴ Total numbers = 11

∴ Prime number are, 11, 13, 17, 19 = 4

And, List the numbers divisible by 3 = 3 numbers (12, 15, 18)

∴ Total favorable outcomes = Primes + Divisible by 3 - Overlap
= 4 + 3 - 0 = 7

∴ Probability = Favorable outcomes/Total outcomes
= 7/11

So the probability is 7/11.

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: At a conference, everyone shakes hands with everybody else. If there were 105 handshakes, how many people were at the conference?

Solution:
ধরি, কনফারেন্সে x সংখ্যক লোক ছিল।
দুজন ব্যক্তির মধ্যে একটি হ্যান্ডশেক হয়, যা Combination দিয়ে প্রকাশ করা যায়।

প্রশ্নমতে,
^xC₂ = 105
⇒ x!/(2!(x - 2)!) = 105
⇒ {x(x - 1)(x - 2)!}/{2 × 1 × (x - 2)!} = 105
⇒ x(x - 1)/2 = 105
⇒ x(x - 1) = 210
⇒ x² - x - 210 = 0
⇒ x² - 15x + 14x - 210 = 0
⇒ x(x - 15) + 14(x - 15) = 0
⇒ (x - 15)(x + 14) = 0

হয় x - 15 = 0 অথবা x + 14 = 0
∴ x = 15 অথবা x = -14

যেহেতু লোকসংখ্যা ঋণাত্মক হতে পারে না, তাই x = 15

অতএব, কনফারেন্সে 15 জন লোক ছিল।

We know, the probability of an event = Favorable outcomes / Total outcomes.
In a case of two dices, Total outcomes = 6 × 6 = 36
A.T.Q favorable outcome is a sum of 6,
Ways of obtaining a sum of 6 (1,5), (5,1), (2,4), (4,2) and (3,3).
Thus, there are 5 ways in which 6 can be obtained using two dices.
Therefore, required probability, P = 5/36

There are 9 letters in the given word, out of which 4 are vowels.
In the word ''CANDIDATE'' we treat the vowels ''AIAE'' as one letter.
Thus, we have CNDDT(AIAE).
Now, we have to arrange 6 letters, out of which D occurs twice.
Therefore,
Number of ways of arranging these letters = 6!/2!
= 720/2
= 360 ways.
Now,
AIAE has 4 letters, in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = (1 × 2 × 3 × 4)/2
= 12
Therefore, required number of words = (360 × 12)
= 4320.

Question: If ⁿC₇ = ⁿC₅ , then find the value of n.

Solution:
We know, 
If ⁿC_r = ⁿC_s , then, n = r + s

Now, ⁿC₇ = ⁿC₅ 
∴ n = 7 + 5
= 12

Let S be the sample space
Then,
n(S) = ⁵²C₂
= 52 × 51/2 × 1
= 1326
Let E = event of getting 2 kings out of 4.
n(E) = 4C2
=> 4 × 3/2 × 1
P(E) = n(E)/n(S)
= 6/1326
= 1/221