ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

Solution:
Total outcomes = {TTT, TTH,THT, HTT, THH, HTH, HHT, HHH} = 8
Favorable outcomes = {HHT, HTH, THH, HHH} = 4

So, the probability of getting at least 2 heads = Favorable outcomes/Total outcomes
= 4/8 = 1/2

Question: ⁷P_r = 210 and ⁷C_r = 35 then what is the value of r?
​
​​​​Solution:
​Given that,
​⁷P_r = 210 and ⁷C_r = 35
​
​We know that,
​ⁿP_r​  = r! × ⁿC_r​
​⇒ 210 = r! × 35
 ​⇒ ​r! = 210/35
​ ​⇒ r! = 6
 ​⇒ ​r! = 3!
∴ ​r = 3

Question: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Solution:
In a simultaneous throw of two dice,
we have n(S) = (6 × 6) = 36

Now, we find the odd product,
E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5 ,5)}

∴ n(E) = 9

∴ P(odd product)= 9/36 = 1/4

​​Now, we find the even product,
​probability(product even) = 1 - (1/4) = 3/4

Question: The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?

Solution:
Here,
The order of each letter in the dictionary is ABLORU.

Now,
with A in the beginning, the remaining letters can be permuted = 5! ways.
= 120 ways

Similarly,
with B in the beginning, the remaining letters can be permuted = 5! ways.
= 120 ways

With L in the beginning,
the first word will be LABORU, the second will be LABOUR.

Hence, the rank of the word LABOUR = 5! + 5! + 2
= 120 + 120 + 2
= 242

Question: Two cards are drawn at random and without replacement from a standard deck of 52 cards. What is the probability that both cards are face cards?

Solution:
Total card = 52
Total face card = 3 × 4 = 12

Total ways to choose 2 cards from 52 = ⁵²C₂ = (52 × 51)/2 = 1326

Total ways to choose 2 face cards from 12 = ¹²C₂ = (12 × 11)/2 = 66

∴ So, the probability that both cards are face cards = 66/1326
= (2 × 3 × 11)/(2 × 3 × 221)
= 11/221

Question:  A group has 8 men and 7 women. In how many ways can a committee of 5 people be formed if the number of women is at least 3?
​
​Solution:
​Given that,
​Total committee size = 5

And for women ≥ 3, possible distributions:
Three ways to formed the committee
1. 3 women + 2 men
2. 4 women + 1 man
3. 5 women + 0 men
​
​Now, 1st case- 3 women + 2 men
​Choose 3 women from 7, (⁷C₃) = 35
​Choose 2 men from 8,  (⁸C₂) = 28
​∴ Total ways = 35 × 28 = 980
​
​2nd case- 4 women + 1 man
​Choose 4 women from 7, (⁷C₄) = 35
Choose 1 man from 8, (⁸C₁) = 8
 ​∴ Total ways = 35 × 8 = 280
​
​And 3rd case-​5 women + 0 men
Choose 5 women from 7,  (⁷C₅) = 21
No men to choose
​∴ Total ways = 21
​
​∴ ​Total ways = 980 + 280 + 21= 1281

Question: Six friends Rita, Anika, Zara, Lima, Arif, and Sayeed sit randomly in a row of six chairs. What is the probability that Rita and Anika do not sit next to each other?

Solution:
Total number of possibilities = 6! = 720

Number of possibilities where Rita and Anika sit together = 5! × 2! 
​= 120 × 2
= 240

So the possibilities where Rita and Anika do not sit together = 720 - 240
= 480

∴Probability that Rita and Anika do not sit next to each other = 480/720
= 2/3

Question: Fifteen distinct points are randomly placed on the circumference of a circle. At most how many triangles can be formed using these points?
​
​Solution:
​Given that,
​Number of distinct points = 15
​Maximum number of triangles = ¹⁵C₃
​= 15!/3!(15 - 3)!
​= (15 × 14 × 13 × 12!)/(3 × 2 × 12!)
= 455
​

Question: A bag contains 6 red, 4 green, and 5 blue balls. Two balls are drawn without replacement. What is the probability that the first ball is red and the second ball is green or blue?
​
​Solution:
​Given that,
​Red balls = 6
​Green balls = 4
​Blue balls = 5
​Total balls = 6 + 4 + 5 = 15

​We know 
​Probability = Favorable outcomes/Total outcomes

​Now,
​First ball is Red = 6/15 = 2/5
​If first is red, then remaining balls = 14
​Second ball is Green or Blue = (4 + 5)/14 = 9/14
​
​∴ Required probability = (2/5) × (9/14) 
​= 9/35
​
​∴ The required probability is 9/35.

Question: How many 8 letter words can be formed by rearranging the letters of the word TRENDING such that T and G occupy the first and last positions respectively?

Solution:
As T and G should occupy the first and last position, the first and last position can be filled in only one following way.
T _ _ _ _ _ _ G.

The remaining 6 positions can be filled in the remaining words (R, E, N, D, I, N) where "N" comes twice.

Total permutations of these 6 letters with one letter repeating = 6!/2! = 720/2 = 360 ways

Question: How many words can be formed by using 3 letters from the word 'DELHI'? 

Solution:
Here we will use the Permutations for this question.
We know,
​ ⁿP_r, for this we have,
n = 5, Total 5 Letters
r = 3, Letters word we required

​Now,
ⁿP_r = n!/(n-r)!
= ⁵P₃ 
​= 5!/2! 
​= 120/2 
​= 60

So, Total we can form 60 different permutation of word from Letter Delhi.

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 white?
​
​​​Solution:
​Given that,
​Number of black balls = 5
​Number of white balls = 6

​Now, 
​Favorable event = ⁶C₂ = 15
​Total possible events = ¹¹C₂ = 55
​
​∴ Probability = 15/55 = 3/11

Question: How many words can be formed by using the letters from the word 'DRIVER' such that all the vowels are never together?

Solution:
We assume all the vowels to be a single character, i.e., 'IE' is a single character.
So, now we have 5 characters in the word, namely, D, R, V, R, and IE.

But, R occurs 2 times.
Number of possible arrangements = 5!/2! = 60

Now, 
​the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together = 60 × 2 = 120

Total number of possible words = 6!/2! = 720/2 = 360

Therefore, the total number of possible words such that the vowels are never together = 360 - 120 = 240

Question: In how many ways can we select a team of 5 students from a given choice of 20?

Solution:
The number of possible ways of selection is given by,
20C5
​= 20!/5!(20 - 5)!
= (20 × 19 × 18 × 17 × 16 × 15!)/(5 × 4 × 3 × 2 × 15!)
​= 15504

So, the number of ways to select 5 students from 20 is 15504.