ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: A dice is thrown randomly. What is the probability that the number shown on the dice is not divisible by 3?

Solution:
Numbers on dice are {1, 2, 3, 4, 5, 6}
Numbers on dice not divisible by 3 are {1, 2, 4, 5}
Number of favorable outcomes = 4
Total possible outcomes = 6

∴ The probability that the number shown on the dice is not divisible by 3 is 4/6 = 2/3

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

Solution:
Total no. of letters in the word 'INSTITUTE' = 9
Repeating letters:
I = 2 times
T = 3 times
∴ Required no. of ways = 9!/(2! × 3!)
= (9 × 8 × 7 × 6 × 5 × 4)/2
= 30240

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

Solution:
The letters of ASSISTANT are A A S S S T T I N
The letters of STATISTICS are A S S S T T T I I C
Common letters are A, S, T, I
The probability of choosing an A from ASSISTANT & an A from STATISTICS is:
(²C₁/⁹C₁) × (¹C₁/¹⁰C₁)
= (2/9) × (1/10)
= 1/45

The probability of choosing an I from ASSISTANT & an I from STATISTICS is:
(¹C₁/⁹C₁) × (²C₁/¹⁰C₁)
= (1/9) × (2/10)
= 1/45

The probability of choosing a S from ASSISTANT & a S from STATISTICS is:
(³C₁/⁹C₁) × (³C₁/¹⁰C₁)
= (3/9) × (3/10)
= 1/10

The probability of choosing a T from ASSISTANT & a T from STATISTICS is:
(²C₁/⁹C₁) × (³C₁/¹⁰C₁)
= (2/9) × (3/10)
= 1/15

∴ The probability that the chosen letters are the same letters is:
1/45 + 1/45 + 1/10 + 1/15
= (2 + 2 + 9 + 6)/90
= 19/90

Question: A bag contains a certain number of bolts out of which some are 'defective' while the remaining are 'non-defective'. The probability of picking a 'non-defective' bolt is 200% more than picking a 'defective' bolt from the bag. If 40 more 'non- defective' bolts are added to the bag then the probability of picking a 'defective' bolt becomes 20%. Now two bolts are picked from the bag then what is the probability that at most one of the two bolts is 'defective'?

Solution:
Let the number of non-defective bolts in the bag = N
The number of defective bolts in the bag = D
∴ The total number of bolts in the bag = N + D

The probability of picking a 'non-defective' bolt is 200% more than picking a 'defective' bolt from the bag.
ATQ,
N/(N + D) = 300% × D/(N + D)
∴ N = 3D ..............(i)

If 40 more 'non-defective' bolts are added to the bag then the probability of picking a 'defective' bolt becomes 20%.
D/(N + D + 40) = 20/100
⇒ D/(3D+ D+ 40) = 1/5 [Using (i)]
⇒ 5D = 4D + 40
∴ D = 40
∴ N = 3D = 120

The total number of bolts in the bag = 40 + 120 = 160

Two bolts are picked from the bag then the probability that at most one of the two bolts is 'defective' = 1 - P (Both the selected bolts are defective)

∴ Required probability = 1 - (⁴⁰C₂/ ¹⁶⁰C₂)
= 1 - (780/12720)
= 1 - (78/1272)
= 1 - (13/212)
= (212 - 13)/212
= 199/212
Hence, the correct answer is 199/212.

Question: In how many different ways can the letters of the word 'BANKING' be arranged so that the vowels always come together?

Solution:
In the word 'BANKING' we treat the two vowels 'A' and 'I' as one letter.
Thus, we have BNKNG(AI)

This has 6 letters of which N occurs 2 times and the rest are different.
As we know that number of ways of arranging n letters out of which r are of same type = n!/r!
Number of ways of arranging these letters = 6!/2! = 360

Now 2 vowels (A,l) can be arranged in 2! = 2 ways

∴ Required number of ways = 360 × 2 = 720

Question: There is a bag that contains 4 yellow balls, 5 black balls and 7 pink balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is black is:

Solution:
Total number of balls = 16
The required number of ways
Case 1:
1 black and 2 others = ⁵C₁ × ¹¹C₂
= 5 × 55
= 275

Case 2:
2 black and 1 other = ⁵C₂ × ¹¹C₁
= 10 × 11
= 110

Case 3:
All the three black = ⁵C₃ = 10

Total = 275 + 110 + 10 = 395 ways
Hence option number (1) is the right answer

Question: How many 4 letter code can be formed using the first 9 letters of the English alphabets, if no letter can be repeated?

Solution:
By fundamental principle, it is (9 × 8 × 7 × 6) ways
= 3024 ways

Alternative:
⁹P₄ = 9!/(9 - 4)!
= 9!/5!
= (9 × 8 × 7 × 6 × 5!)/5!
= 3024 ways

Question: A box contains three colored marbles, such that the number of blue, yellow and grey marbles is 8, 7 and 5 respectively. Find the probability of picking two marbles, one being blue marble and the other being grey marble.

Solution:
From the given data,
Number of blue marbles = 8
Number of yellow marbles = 7
Number of grey marbles = 5

∴ Probability = (⁸C₁ × ⁵C₁)/²⁰C₂
= (8 × 5)/190
= 40/190
= 4/19.
Hence, option(4) is correct.

Question: In how many ways can 3 people be seated in a row containing 6 seats?

Solution:
First person can be seated in 6 ways, the second person in 5 ways and the third person in 4 ways.
Then, by fundamental principle, total number of ways in which three persons can be seated in 6 seats in a row is (6 × 5 × 4) ways
= 120 ways

Question: In how many ways can seven books be arranged on a shelf?

Solution:
Number of ways in which the first book can be placed = 7
Number of ways in which the second book can be placed = 6
Similarly,
The total number of ways in which seven books can be arranged on a shelf = 7 × 6 × 5 × 4 × 3 × 2 × 1
= 7!
= 5040

Question: Probability of 3 students solving a question are 1/2, 1/3, and 1/4. Probability to solve the question is:

Solution:
Probability of 3 students,
P(A) = 1/2, ∴ P(A′) = 1/2
P(B) = 1/3, ∴ P(B′) = 2/3
P(C) = 1/4, ∴ P(C′) = 3/4

So, Probability of no one solve the question is = (1/2) × (2/3) × (3/4)
= 1/4
∴ P(None) = 1/4

Then, The probability to solve the question is = 1 - 1/4
= 3/4
Hence, the correct answer is 3/4.

Question: Find the number of ways of selecting 4 girls and 4 boys out of 7 girls and 6 boys.

Solution:
The number of ways of selecting 4 girls and 4 boys out of 7 girls and 6 boys is
⁷C₄ × ⁶C₄
= {7!/(4! × 3!)} × {6!/(4! × 2!)}
= (35 × 15)
= 525 ways

Question: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7 if no digit is repeated?

Solution:
Given digits: 1, 2, 3, 4, 6, 7
Number of digits = 6

Number of possible digits at unit’s place = 3 (2, 4 and 6)
⇒ Number of permutations = ³P₁ = 3

When one of the digits is taken in units’ place, then the number of possible digits available = 5
⇒ Number of permutations = ⁵P₂ = 5!/(5 - 2)!
= 5!/3!
= 120/6
= 20

∴ The total number of permutations = 3 × 20 = 60.