ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

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  black?

Solution: 
Given that 
Number of black balls = 5
Number of white balls = 6
Favorable event = ⁵C₂
Total possible events = ¹¹C₂
∴ Probability = ⁵C₂/¹¹C₂
= 10/55
= 2/11

Question: A bag contains 8 red and 5 green balls. Two balls are drawn randomly. Find the probability that one ball is red and other is green.

Solution: 
A bag contains 8 red and 5 green balls. 
Total balls = 8 + 5 = 13
Two balls drawn ¹³C₂ = 78
One ball is red and other is green = ⁸C₁ × (⁵C₁) = 8 × 5 = 40

Required probability = 40/78 = 20/39

Question: In how many ways can the letters of the word 'CANVAS' be arranged?

Solution: 
The word 'CANVAS' contains 6 letters, namely 1C, 2A, 1N, 1V and 1S


Required number of ways =6!/2! = 360

Question: In a single throw of a die, what is the probability of getting a number greater than 4?

Solution: 
In a single throw of dice, total results are 1, 2, 3, 4, 5 and 6 = 6 results
Event of getting numbers greater than 4 ={5, 6} = 2 possible results
Required probability to get a number greater than 4 is 2/6 = 1/3 

Question: In how many different ways can the letters of the word 'RIDDLED' be arranged? 

Solution: 
The given word contains 7 letters of which D is taken 3 times.
∴ Required number of ways = 7!/3!
= 840

Question: If 5 × ⁿP₃ = 4 × ^{(n + 1)}P₃, find n? 

Solution: 
5 × ⁿP₃ = 4 × ^{(n + 1)}P₃
5 × n × (n - 1) × (n - 2) = 4 × (n + 1) × n × (n - 1)
Or, 5(n - 2) = 4(n + 1)
Or, 5n - 10 = 4n + 4
Or, 5n - 4n = 4 + 10
Hence, n = 14

Question: In a box, there are 8 red, 7 blue and 5 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor red?

Solution: 
Total number of balls = (8 + 7 + 5) = 20.

Let E = event that the ball drawn is neither blue nor red
         = event that the ball drawn is green.
n(E) = 5
P(E) =n(E)/n(S) = 5/20 = 1/4
       

Question: In how many ways 2 students can be chosen from the class of 18 students?

Solution: 
Number of ways =¹⁸C₂ =  153

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

Solution: 
Clearly, 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 how may different ways can the letters of the word PIRATE be arranged in such a way that the vowels always come together? 

Solution: 
The word 'PIRATE' contains 6 different letters.
When the vowels IAE are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PRT(IAE).
Now, 4 letters can be arranged in 4! ways 
                                                   = 24 ways.
The vowels (IAE) can be arranged among themselves in 3! = 6 ways.

∴ Required number of ways = (24 x 6) = 144

Question: In a simultaneous throw of two coins, the probability of getting at least one head is: 

Solution: 
Here S= {HH,HT,TH,TT}
Let E= event of getting at least on head ={HH,HT,TH}

∴P(E)= n(E)/n(S)​ = 3/​4

Question: If ¹⁸C_r = ¹⁸C_{r + 2} ,find ^rC₅ = ?

Solution: 
¹⁸C_r ​= ¹⁸C_{r + 2​}
So, r + r + 2  = 18
2r + 2 = 18 
2r = 18 - 2
2r = 16
r = 8 
 
⁸C₅ = 56

Question: A box contains 4 red, 5 green and 6 black balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or black?

Solution: 
Total number of balls = (4 + 5 + 6) = 15
P(drawing a red ball or a black ball) = P(red) + P(black)
                                                          = (4/15) + (6/15)
                                                          = (4 + 6)/15
                                                          = 10/15
                                                          = 2/3
∴ The probability is 2/3.

Question: A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 7 from part A and 6 from part B, in how many ways can he choose the questions?

Solution: 
No. of questions in part A = 10
No. of questions in part B = 10

A student has to choose 8 from part A and 6 from part B

Required number of ways = (¹⁰C₇​ × ¹⁰C₆​) =120  × 210 = 25,200 

Question: How many different four digit numbers can be formed using the digits 1,2,7,4,5,6 when repetition is not allowed and each number starts with 2?

Solution: 
Here
1,2,7,4,5,6

No repetition Four digit number Starting with 2 
Rest leftover numbers = 5 
Number of blanks = 3 
Number of digits 
⁵P_3​
= 5 × 4 × 3
= 60