ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: Two dice are tossed. The probability that the total score is a prime number is:

Solution:
n(S) = (6 × 6) = 36.
Let E = Event that the sum is a prime number.
Then E= {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5)}
n(E) = 15.

∴ P(E) = n(E)/n(S)
= 15/36
= 5/12

Question: In a simultaneous throw of pair of dice. Find the probability of getting the total more than 7.

Solution:
Here,
n(S) = (6 × 6) = 36

Let E = event of getting a total more than 7
= {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Therefore,P(E) = n(E)/n(S)
= 15/36
= 5/12

Question: How many different ways can the letters in the word ATTEND be arranged?

Solution:
There are 6 letter in the word 'ATTEND' whereas, T comes two times.
So, required number of ways = 6!/2!
= 360

Question: A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

Solution:
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7 =⁷C₂ = 21
Let E = Event of drawing 2 balls, none of which is blue.
n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls = ⁵C₂ = 10
Therefore, P(E) = n(E)/n(S)
= 10/ 21

Question: From a group of 6 men and 4 women a Committee of 4 persons is to be formed. In how many different ways can it be done, so that the committee has at least 2 men?

Solution:
The committee of 4 persons is to be so formed that it has at least 2 men. The different ways that we can choose to form such a committee are:
2m. 2w in = ⁶C₂ × ⁴C₂ = 90
3m. 1w in = ⁶C₃ × ⁴C₁ = 80
4m in = ⁶C₄ = 15

∴ Total no. of different ways in which a committee of 4 persons can be formed so that it has at least 2 men = 90 + 18 + 15
= 185

Question: In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?

Solution:
No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!

Question: In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Solution:
probability of getting a prize = 10/(10 + 25)
= 10/35
= 2/7

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

Solution:
Probability = ⁵C₁ × ³C₁ - 1
= 15 - 1
= 14

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

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

∴ P(E) = n(E)/n(s)
= 2/52
= 1/26

Question: A word consists of 9 letters; 5 consonants and 4 vowels.Three letters are choosen at random. What is the probability that more than one vowel will be selected?

Solution:
3 letters can be choosen out of 9 letters in ⁹C₃ ways.
More than one vowels ( 2 vowels + 1 consonant  or  3 vowels ) can be choosen in (⁴C₂ × ⁵C₁) + ⁴C₃ ways

Hence,required probability = {(⁴C₂ × ⁵C₁) + ⁴C₃}/⁹C₃
= 17/42

Question: Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?

Solution:
There are 6 candidates and a voter has to vote for any two of them.
So, the required number of ways is = ⁶C₂
= 15

Question: There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

Solution:
Total cases of checking in the hotels = 4 × 4 × 4 = 64 ways.
Cases when 3 men are checking in different hotels = 4 × 3 × 2 = 24 ways.

Required probability = 24/64
= 3/8

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

Question: A question paper consists of three sections 4,5 and 6 questions respectively. Attempting one question from each section is compulsory but a candidate need not attempt all the questions. In how many ways can a candidate attempt the questions?

Solution:
At least 1 question from each section is compulsory, so from the 1st section the candidate can attempt 1 or 2 or 3 or 4 questions.
In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it.
So far 4 question there are 2 × 2 × 2 × 2 ways to attempt.
As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 2⁴ - 1
Similarly for the 2nd section there are 2⁵ - 1 ways in which he can attempt and for the 3rd section there are 2⁶ - 1 ways.
The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections.
Thus, total number of ways in which he can attempt questions in that paper:
= (2⁴ - 1)(2⁵ - 1)(2⁶ - 1)
= 15 × 31 × 63
= 29295

Question: Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.

Solution:
The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.
So the required probability = {(8/17) × (9/16)} + {(9/17) × (8/16)} + {(8/17) × (7/16)}
= (9/34) + (9/34) + (7/34)
= (9 + 9 + 7)/34
= 25/34

Question: There are 10 oranges in a basket. Find the no. of ways in which 3 oranges are chosen from the basket?

Solution:
Required number of ways = ¹⁰C₃
= 10!/{3! (10 - 3)!}
= 10!/(3! 7!)
= (10 × 9 × 8)/(3 × 2)
= 120

Question: In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Solution:
Total number of balls
= (8 + 7 + 6)
= 21

Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue
= 7/21
= 1/3

Question: In a hockey championship, there are 153 matches played. Every two team played one match with each other. The number of teams participating in the championship is:

Solution:
Let there were x teams participating in the games, then total number of matches,
nC2 = 153
⇒ {n × (n - 1)}/2 = 153
⇒ n(n - 1) = 306
⇒ n² - n - 306 = 0

Since n typically represents a count or a positive quantity in such contexts, we discard the negative solution.
Therefore, n = 18.

Question: What is the probability of getting 53 Mondays in a leap year?

Solution:
1 year = 365 days . A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 × 7 = 364days
366 - 364 = 2 days

In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
1. Sunday, Monday
2. Monday, Tuesday
3. Tuesday, Wednesday
4. Wednesday, Thursday
5. Thursday, Friday
6. Friday, Saturday
7. Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.

Hence the probability of getting 53 days = 2/7

Question: Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.

Solution:
Let A be the event that X is selected and B is the event that Y is selected.
P(A) = 1/7,
P(B) = 2/9.

Let C be the event that both are selected.
P(C) = P(A) × P(B) as A and B are independent events:
= (1/7) × (2/9)
= 2/63