ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: If the probability of rain on any given day in City Dhaka is 50 percent, what is the probability that it rains on exactly 2 days in a 4-day period?

Solution:
If the probability of rain on any given day in City Dhaka is 50 percent
the probability of rain on any given day = 1/2
the probability of no rain on any given day = 1/2

selecting 2 days out of 4 = ⁴C₂

∴the probability that it rains on exactly 2 days in a 4-day period is = ⁴C₂ × 1/2 × 1/2 × 1/2 × 1/2
= 6 × 1/2⁴
= 6 × 1/16
= 3/8

Question: In how many ways, a committee of 5 members be selected from 7 men and 5 ladies, consisting of 3 men and 2 ladies?

Solution:
there are total 7 men and 5 ladies

∴ number of ways a committee of 5 members can be slected = (⁷C₃) × (⁵C₂)
= 35 × 10
= 350 ways

Question: If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 4 times, what is the probability that it will land heads up on the first 3 flips and not on the last flip?

Solution: 
The probability of landing heads and not landing on heads is same = 1/2
The probability of first three heads =(1/2) × (1/2) × (1/2)
The probability of last  landing not on heads = 1/2
The total probability =(1/2) × (1/2) × (1/2) × (1/2)
= 1/ 2⁴
= 1/16

Question: How many 3-digit numbers can be formed from the digits 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Solution:
Number will be divisible by 5 if the last number is 5.
So, first number can be chosen in ⁴C₁ ways 
= 4 ways

As the digit is not repeated
second number can be chosen in ³C₁ 
= 3 ways

∴ Total ways = 4 × 3 ways
= 12 ways

Question: A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 3 women?

Solution:
A small company employs 3 men and 5 women.
Total people = 8

ways of selecting 4 people from 8 = ⁸C₄
= 70

ways of selecting 3 women from 5 = ⁵C₃
ways of selecting 1 men from 3 = ³C₁

∴ probability = (⁵C₃ × ³C₁)/ 70
= (10 × 3)/70
= 3/7

Question: An integer n between 1 and 100, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 5?

Solution: 
total number = 100
n(n+1) will be divisible by 5 if n or n + 1 is divisible by 5

when n is divisible by 5, there are 20 such number (5, 10, 15, 20, 25,.....,100)
when n + 1 is divisible by 5, there are 20 such number (4, 9, 14,.....,99)

∴ proability = (20 + 20)/100
= 40/100
= 2/5

Question: Five persons, Shahin, Masud, Jayed, Raj, and Alom, sit randomly in five chairs in a row. What is the probability that Raj and Masud sit next to each other?

Solution:
Total possibilities = 5! = 120
favorabole events = 4! × 2! 
= 24 × 2
= 48

probability = 48/120
= 2/5

Question: In how many ways can 5 examination papers be arranged so that the best and the worst papers never come together?

Solution:
Total ways = 5!
= 120 ways

if two papers come together, we can consider them one.
ways that they will come together = 4! × 2!
= 24 × 2
= 48 ways

∴ ways the best and the worst papers never come together = 120 - 48 ways
= 72 ways

Question: If x and y are two positive integers and x + y = 4 then, what is the probability of x equals to 1?

Solution:
total possible ways = (1, 3), (2, 2), (3, 1) = 3
favorable event = (1, 3) = 1

∴ probability = 1/3

Question: In how many ways 5 students can be chosen from the class of 10 students?

Solution:
ways 5 students can be chosen from the class of 10 students is = ¹⁰C₅
= 10!/(5! 5!)
= 252

Question: The ratio of red balls, to yellow balls, to green balls in a basket is 2 : 3 : 4. What is the probability that a ball chosen at random from the basket is a yellow ball?

Solution:
The ratio of red balls, to yellow balls, to green balls in a basket is 2 : 3 : 4
let, there are 2x red balls, 3x yellow balls and 4x green balls 
total balls = 2x + 3x + 4x
= 9x

∴ probability that a ball chosen at random from the basket is a yellow ball = 3x/9x
= 3/9
= 1/3

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

Solution:
ways  one can cast his vote = ⁵C₂
= 5!/2! 3!
= 10

Question: A prize of 2000 tk is given to anyone who solves a hacker puzzle independently. The probability that Rafi will win the prize is 0.6, and the probability that kamol will win the prize is 0.7. What is the probability that both will win the prize?

Solution:
probability that Rafi will win the prize is, P(A) = 0.6
probability that kamol will win the prize is, P(B) = 0.7
As events are independent,
P (A ∩ B) = P(A) × P (B)
= 0.6 × 0.7
= 0.42

Question: A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 5 from part A and 8 from part B, in how many ways can he choose the questions?

Solution:
ways to choose 5 from part A = ¹⁰C₅
ways to choose 8 from part B = ¹⁰C₈

choose 5 from part A and 8 from part B = ¹⁰C₅ × ¹⁰C₈
= {10!/(5! 5!)} × {10!/(2! 8!)}
= 11340

Question: According to meteorological records, it rained on 21 days in the month of June last year. What is the probability that it will rain on fourth of June this year?

Solution:
June month has 30 days
favorable events = 21 days

∴ the probability that it will rain on fourth of June this year = 21/30
= 7/10

Question: In a survey among the readers of newspapers, it was found that 65 persons read the Prothom Alo, 40 persons read the Bhorer Kagoj, 45 read the Janakantho, 52 read the Jugantor. If one person is chosen at random from the readers, what is the probability that the person doesn't read the Jugantor? 

Solution:
Total persons = 65 + 40 + 45 + 52
= 202

probability of that person reading jugantor = 52/202
= 26/101

the probability that the person doesn't read the Jugantor = 1 - 26/101
= (101 - 26)/101
= 75/101

Question: In a party every person shakes hands with every other person. If there are 55 hands shakes, find the number of person in the party.

Solution:
Let n be the number of persons in the party
Total number of hands shake is given by ⁿC₂

so,
ⁿC₂ = 55
⇒ n!/2! (n - 2)! = 55
⇒ n(n - 1)/2 = 55
⇒ n2 - n = 110
⇒ n2 - n - 110 = 0
⇒ n2 - 11n + 10n - 110 = 0
⇒ n(n - 11) + 10(n - 11) = 0
⇒ (n - 11) (n + 10) = 0

∴ n + 10 = 0
n = - 10, not possible
∴ n - 11 = 0
n = 11

so, there are 11 persons.

Question: If the probability that Mokhles will miss at least one of the ten jobs assigned to him is 0.55, then what is the probability that he will do all ten jobs?

Solution:
There are only two cases:
1) Mokhles will miss at least one of the ten jobs.
2) Mokhles will not miss any of the ten jobs.

Hence, (The probability that Mokhles will miss at least one of the ten jobs) + (The probability that he will not miss any job) = 1. Since the probability that Mokhles will miss at least one of the ten jobs is 0.55, this equation becomes
⇒ 0.55+ (The probability that he will not miss any job) = 1
⇒ (The probability that he will not miss any job) = 1 - 0.55
⇒ (The probability that he will not miss any job) = 0.45

Question: If, 4 × ⁿP₃ = 3 × ^{(n + 1)}P₃, what is the value of n?

Solution:
4n!/(n - 3)! = 3(n +1)!/(n + 1 - 3)!
⇒ 4 n(n - 1)(n - 2)(n - 3)!/(n - 3)! = 3 (n + 1) n (n - 1) (n - 2)!/(n - 2)!
⇒ 4 n(n - 1)(n - 2) =  3 (n + 1) n (n - 1)
⇒ 4 (n - 2) = 3 (n + 1)
⇒ 4n - 8 = 3n + 3
⇒ 4n - 3n = 3 + 8
∴ n = 11

Question: A meeting is attended by 750 students. 450 of the students are females. Half the female students are less than thirty years old, and one-fourth of the male students are less than thirty years old. If one of the students of the meeting is selected at random to receive a prize, what is the probability that the person selected is not less than thirty years old?

Solution:
 A meeting is attended by 750 students. 450 of the students are females. 

Half the female students are less than thirty years old
number of females less than thirty years old = 450/2
= 225

male students = 750 - 450
= 300
one-fourth of the male students are less than thirty years old.
number of males less than thirty years old = 300/4
= 75

total number of students less than thirty age = 225 + 75
= 300

the probability that the person selected is less than thirty years old = 300/750
= 2/5

∴  the probability that the person selected is not less than thirty years old = 1 - (2/5)
= (5 - 2)/5
= 3/5

Question: ⁶P_m = 360, ⁶C_m = 15, what is the value of m?

Solution:
⁶P_m = 360
⇒ 6!/(6 - m)! = 360........(1)

⁶C_m = 15
⇒ 6!/m! (6 - m)! = 15..........(2)

(1) ÷ (2) ,
{6!/(6 - m)!} / {6!/m! (6 - m)! } = 360/15
⇒ m! = 24
= 4 × 3 × 2 × 1

∴ m = 4

Question: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one girl should be there?

Solution: 
There are total 10 children

Selecting 4 children out of 10 = ¹⁰C₄
= 210 ways

selecting 4 boys out of 6 boys = ⁶C₄
= 15

∴ different ways can they be selected such that at least one girl should be there is = 210 - 15 ways
= 195 ways