Probability, Permutation and Combination

Question: A company offers a bonus to employees who complete certain training modules. The probability that Emma will complete the "Leadership Skills" module is 0.8, and the probability that John will complete the "Time Management" module is 0.5. What is the probability that both Emma and John will complete their respective modules?

Solution:
Let,
Probability that Emma completes the "Leadership Skills" module P(E) = 0.8
P(J) = Probability that John completes the "Time Management" module P(J) = 0.5

Since the events are independent, the probability that both Emma and John will complete their respective modules,
P (E ∩ J) = P(E) × P (J)
= 0.8 × 0.5
= 0.4

Question: A bag contains 5 black and 6 white balls; two balls are drawn at random. What is the probability that the balls drawn are white?
​
​​​Solution:
​Given that,
​Number of black balls = 5
​Number of white balls = 6

​Now, 
​Favorable event = ⁶C₂ = 15
​Total possible events = ¹¹C₂ = 55
​
​∴ Probability = 15/55 = 3/11

Here, one probability is to find which bag is selected AND the other is for a white napkin from the selected bag.
Probability of selecting 1 bag out of 2 bags = 1/2

Say it has 4 white and 2 black napkins.

So, white napkins probability = 4/(4 + 2) = 4/6
So Probability 1 = (1/2) × (4/6) = 4/12

Similarly, Probability 2 = (1/2) × {5/(5 + 3)} = 5/16

Total probability of white napkin = (4/12) + (5/16) = 31/48

In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36
Let E = event of getting a total of 10 or 11
= [(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)]
∴P(E) = n(E)/n(S) = 5/36

Question: In a class, there are 12 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is-

Solution: 

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 22
= ²²C₃ 
= (22 x 21 x 20)/(3 x 2 x 1)
= 1540

n(E) = ¹⁰C₁ x ¹²C₂
= (10 x 12 x 11)/(2 x 1)
= 660

∴ P(E) = n(E)/n(S)
= 660/1540
= 3/7   

Question: A box contains 20 electric bulbs, out of which 4 are defective. Two balls are chosen at random from this box. The probability that at least one of them is defective, is

Solution:
Given that,
Total bulbs = 20
Defective bulbs = 4
Non-defective bulbs = 20 - 4 = 16
Two bulbs are chosen at random (without replacement)

Now,
P(both non-defective) = (16/20) × (15/19) = 240/380 = 12/19

And,
∴ P(at least one defective) = 1 - P(both non-defective)
= 1 - (12/19)
= (19 - 12)/19
= 7/19
∴ The probability that at least one of them is defective is 7/19

Question: Tickets numbered 1 to 30 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 5?

Solution:
Here, S = {1, 2, 3, 4, ...., 29, 30}
n(S) = 30

Let E = event of getting a multiple of 4 or 5
∴ n(E) = {4, 5, 8, 10, 12, 15, 16, 20, 24, 25, 28, 30}

∴ P(E) = n(E)/n(S)
= 12/30
= 2/5

Question: In a badminton tournament, every player plays against every other player exactly once. If there were 66 matches played in total, how many players participated in the tournament?

Solution:
ধরি, টুর্নামেন্টে মোট খেলোয়াড়ের সংখ্যা = x

যেহেতু প্রতিটি খেলোয়াড় অন্য সকল খেলোয়াড়ের সাথে একবার করে ম্যাচ খেলে, তাই মোট ম্যাচের সংখ্যা হবে:
^xC₂ = 66
⇒ x!/{2!(x - 2)!} = 66
⇒ x(x - 1)/2 = 66
⇒ x(x - 1) = 132
⇒ x² - x - 132 = 0
⇒ x² - 12x + 11x - 132 = 0
⇒ x(x - 12) + 11(x - 12) = 0
⇒ (x - 12)(x + 11) = 0
⇒ x = 12 অথবা x = -11

যেহেতু খেলোয়াড়ের সংখ্যা ঋণাত্মক হতে পারে না, তাই x = 12

∴ টুর্নামেন্টে মোট খেলোয়াড়ের সংখ্যা = 12 জন।

যে কোন করমর্দন অথবা কোলাকুলির অংকে শুধু কত জন করমর্দন (Handshake), বা কোলাকুলি করল তা দেয়া থাকবে।
এক্ষেত্রে মনে রাখতে হবে যে প্রত্যেক বার করমর্দন বা কোলাকুলি করার সময় মোট ২ জন লোকের প্রয়োজন।
তাই এক্ষেত্রে সূত্রটি হবে ⁿC₂ = ^{মোট লোক}C_{২ জন সব সময়}
¹⁰C₂ = 10!/2!(10 - 2)!
= 10!/2!8!
= (10 × 9)/2
= 5 × 9
= 45.

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: Eight points are situated on a plane, 4 of them in a straight line, the other 4 elsewhere. How many triangles can be formed by joining 3 points at a time?

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

Solution:
'MISSISSIPPI' শব্দটিতে মোট 11টি বর্ণ রয়েছে।

শর্ত: প্রথম স্থানে 'M' স্থির। এখন বাকি 11 - 1 = 10টি স্থানে বাকি বর্ণগুলোকে সাজাতে হবে।

বাকি 10টি বর্ণের মধ্যে পুনরাবৃত্ত অক্ষর: I (4 বার), S (4 বার), P (2 বার)।

∴ বাকি 10টি বর্ণের বিন্যাস সংখ্যা = 10!/(4! × 4! × 2!)
= 3,628,800/(24 × 24 × 2)
= 3,628,800/1,152
= 3,150

Question: In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.

Solution: 
Let the Arrangement be,
B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways

∴ Required number of ways = 4! × 3!
= 24 × 6
= 144

Question: A jar contains 5 blue, 8 green, and 6 orange marbles. If a marble is picked at random, what is the probability that the marble drawn will be either blue or green?

Solution:
Total number of marbels = 5 + 8 + 6 = 19

If one marbel is picked at random.

P(blue) = 5/19
P(green) = 8/19

P(blue or green) = 5/19 + 8/19
= (5 + 8)/19
= 13/19

Question: In how many ways can the letters of the word 'TRIANGLE' be arranged, if the vowels always stay together?

Solution: 
Triangle শব্দটিতে মোট বর্ণ  8 টি (সবই ভিন্ন) এবং স্বরবর্ণ 3 টি।
যেহেতু, স্বরবর্ণ 3 টি একত্রে থাকবে, সেহেতু স্বরবর্ণ তিনটিকে একটি বর্ণ ধরে মোট বর্ণ সংখ্যা 6 টি। 
এই 6 টি বর্ণকে নিজেদের মধ্যে সাজানো যায় = 6! উপায়ে
আবার, স্বরবর্ণ  3 টিকে নিজেদের মধ্যে সাজানো যায় = 3! উপায়ে

∴ সর্বমোট বিন্যাস = 6! × 3!
= 4320

Question: A fair die is thrown once. What is the probability of getting a prime number?

Solution:
A standard fair die has 6 faces numbered: 1, 2, 3, 4, 5, 6.
The prime numbers from 1 to 6 are: 2, 3, and 5.

Probability = (Number of favorable outcomes)/(Total number of outcomes)
= 3/6
= 1/2

Question: Labib wants to arrange four out of his five saplings in a row on a shelf. If each sapling is in a pot of a different color, in how many different ways can he arrange the four saplings?

Solution:
যেহেতু প্রতিটি চারাগাছ ভিন্ন ভিন্ন রঙের পাত্রে আছে এবং সেগুলোকে একটি সারিতে সাজাতে হবে, তাই এটি একটি বিন্যাসের (Permutation) সমস্যা।

∴ পাঁচটি চারাগাছ হতে চারটি নিয়ে সাজানো যায় = ⁵P₄ উপায়ে
= 5!/(5 - 4)! উপায়ে
= 5! উপায়ে
= 5 × 4 × 3 × 2 × 1 উপায়ে
= 120 উপায়ে

∴ লাবিব মোট 120টি ভিন্ন উপায়ে চারাগাছগুলো সাজাতে পারবে।

মোট বল রয়েছে = 2 + 3 + 2 = 7 টি
নীল বাদে বল আছে = 7 - 2 = 5 টি
∴ বলটি নীল না হবার সম্ভাবনা = ⁵c₂ / ⁷c₂ = 10/21

Total number of students = 20.
Let S be the sample space.
Then, n(S) = number of ways of three scored first mark
n(S) = ²⁰C₃
= 20 x 19 x 18 / 2 x 3
= 20 x 19 x 3

Let,
E be the event of 1 girl and 2 boys.
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12.
n(E) = ⁸C₁ x ¹²C₂
= 8 x 12 x 11 / 1 x 2
= 8 x 6 x 11.

Now, the required probability = n(E)/n(S)
= (8 x 6 x 11)/(20 x 19 x 3)
= 44/95.

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

Solution:
The events when three unbiased coins are tossed = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Total number of events 8

The events of getting at least two heads {HHH, HHT, HTH, THH}

Number of expected events = 4

∴ The probability of getting at least two heads is = 4/8 = 1/2

‘LOGARITHM’ contains 10 different letters.

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

Question: Twelve distinct points are randomly placed on the circumference of a circle. At most how many triangles can be formed using these points?

Solution:
Given that,
Number of distinct points, n = 12
To form a triangle, we need to select 3 points out of n points.

Maximum number of triangles = ¹²C₃
= 12!/{3!(12 - 3)!}
= (12 × 11 × 10 × 9!)/(3 × 2 × 1 × 9!)
= (12 × 11 × 10)/6
= 220

Question: А bоx contains 3 blue, 2 white, and 4 red marbles. If one marble is drawn at random, what is the probability that it will not be a white marble?

Solution:
Given that,
Blue marbles = 3
White marbles = 2
Red marbles = 4

∴ Total marbles = 3 + 2 + 4 = 9
And, number of non-white marbles = Blue + Red = 3 + 4 = 7

We know,
P(not white) = favorable outcomes​/total outcomes
= 7/9​ 

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: A box contains 400 marbles, of which 40% are blue. You pick some marbles of which 25% are blue. Of the remaining marbles, 50% are blue marbles. How many marbles did you pick?

Solution:
Given that, 
Total marbles = 400
Blue marbles initially = 40% of 400 = (40/100) × 400 = 160
Non-blue marbles = 400 - 160 = 240

Let the number picked be x.
∴ Blue picked = 25% of x = 0.25x
and non-blue picked = 0.75x

And, 
Remaining marbles = 400 - x
∴ Remaining blue marbles = 160 - 0.25x

ATQ, Of the remaining marbles, 50% are blue then we get, 
160 - 0.25x = 0.5(400 - x)
⇒ 160 - 0.25x = 200 - 0.5x
⇒ 0.5x - 0.25x = 200 - 160
⇒ 0.25x = 40
⇒ x = 40/0.25
∴ x = 160

So the number of marbles you picked is 160.

Question: In a simultaneous throw of a pair of dice, what is the probability of getting a total more than 8?

Solution: 
The total number of possible outcomes for the pair of dice is the product of the outcomes for each die is-
= 6 × 6 = 36

And,
The favorable outcomes are the combinations where the sum of the two dice is more than 8. These sums can be 9, 10, 11, or 12. The combinations for each sum are-
Sum of 9: (3, 6), (4, 5), (5, 4), (6, 3) - 4 outcomes
Sum of 10: (4, 6), (5, 5), (6, 4) - 3 outcomes
Sum of 11: (5, 6), (6, 5) - 2 outcomes
Sum of 12: (6, 6) - 1 outcome

∴ The total number of favorable outcomes is the sum of these outcomes = 4 + 3 + 2 + 1 = 10

∴ P(sum > 8) = Favorable Outcomes/Total Outcomes = 10/36 = 5/18

Question: If ⁿC₁₀ = ⁿC₈ , 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,
ⁿC₁₀ = ⁿC₈
⇒ 10 + 8 = n
⇒ n = 18

∴ ⁿC₂ = ¹⁸C₂
= 18!/(2!(18 - 2)!)
= 18!/(2! × 16!)
= (18 × 17 × 16!)/(2 × 1 × 16!)
= (18 × 17)/2
= 9 × 17
= 153

As, Mr. x is always chosen, then to form a committee of 5 members from the pool of 8, we now have to choose 4 members from 7 people

So, the committee can be formed in ⁷c₄ = 35 ways

Question: A committee of 5 members is to be selected from 7 men and 4 women. In how many ways can this be done if exactly 3 men must be selected?

Solution:
এখানে, একটি কমিটি গঠন করতে হলে 3 জন পুরুষ এবং (5 - 3) = 2 জন মহিলা নির্বাচন করতে হবে।

7 জন পুরুষ থেকে 3 জন পুরুষ নির্বাচন করার উপায়:
⁷C₃ = 7!/{3! (7 - 3)!}
= (7 × 6 × 5)/(3 × 2 × 1)
= 35 টি

4 জন মহিলা থেকে 2 জন মহিলা নির্বাচন করার উপায়:
⁴C₂ = 4!/{2! (4 - 2)!}
= (4 × 3)/(2 × 1)
= 6 টি
সুতরাং, মোট সম্ভাব্য উপায় = 35 × 6 = 210 টি।

অতএব, কমিটি গঠনের মোট উপায় হলো 210 টি।

Question: A box contains 5 green, 3 yellow, and 4 black balls. If one ball is drawn at random, what is the probability that it will not be a green ball?

Solution:
Given that,
Green balls = 5
Yellow balls = 3
Black balls = 4
∴ Total balls = 5 + 3 + 4 = 12

And, number of non-green balls = Yellow + Black = 3 + 4 = 7

We know,
P(not green) = favorable outcomes/total outcomes
= 7/12

∴ The probability of drawing a non-green ball is 7/12

Question: A hat contains a total of 30 cards, of which 12 are marked with a star and the remaining 18 are unmarked. If a card is drawn at random from the hat, what is the probability that it will be a card marked with a star?

Solution:

Total number of cards = 12 (marked) + 18 (unmarked) = 30

Number of favorable outcomes (marked with a star) = 12

Probability = (Number of favorable outcomes)/(Total number of outcomes)
= 12/30
= 2/5

Question: Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Solution:
Here, S = {1, 2, 3, 4, ...., 19, 20}
Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}

∴ P(E) = n(E)/n(S)
= 9/20

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= ⁷C₃× ⁴C₂
= {(7 × 6 × 5)/(3 × 2 × 1)} × {(4 × 3)/(2 × 1)}
= 210.

Question: Three unbiased coins are tossed simultaneously. What is the probability of getting at most one tail?

Solution: Total outcomes = 2³ = 8 (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)

Favorable outcomes = cases with at most one tail (zero tails or exactly one tail) = (HHH, HHT, HTH, THH)

At most one tail refers to zero or one tail only.

Therefore, Probability = Favorable outcomes / Total outcomes = 4/8 = 1/2