ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: A letter is taken out at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

Solution: 
For S = (3/9) × (3/10) = 1/10
For A = (2/9) × (1/10) = 1/45
For I = (1/9) × (2/10) = 1/45
For T = (2/9) × (3/10) = 1/15

Total probability = (1/10) + (1/45) + (1/45) + (1/15)
= 19/90 

Question: A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red is :

Solution: 
There are 2 possible cases: the first ball drawn is red or black.
P(first ball black) = 6/10
P(first ball red) = 4/10

P(second ball red | first ball drawn is black) = 4/12
P(second ball red | first ball drawn is red) = 6/12 = 1/2

The probability of the second ball being red, P(R) = (4/10)×(1/2) + (6/10) ×(4/12)
= (1/5) + (1/5)
= 2/5

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: 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: A seven-digit number is formed using the digit 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is:

Solution: 
শেষের অঙ্ক ৪ রেখে বিন্যাস = 6!/(2!2!2!)
= 90 

মোট বিন্যাস = 7!/2!3!2! = 210

সম্ভাব্যতা = 90/210
= 3/7

Question: How many numbers are there between 100 and 1000 inclusive, having at least one of their digits 7?

Solution:
total numbers = 901

3 digit number without 7 = (8 × 9 × 9)
= 648
so, numbers are there between 100 and 1000 inclusive without 7 is = 648 + 1 = 649

∴ numbers are there between 100 and 1000 inclusive, having at least one of their digits 7 = 901 - 649
= 252

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: 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: 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: 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