ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

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

Solution:
S = {1, 2, 3, 4, 5, 6}
n(S) = 6

Then,
E(not divisible by 3) = {1, 2, 4, 5}
n(E) = 4

∴ P(not divisible by 3) = 4/6
= 2/3

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

Solution:
CYCLE whereas total 5 letters and C comes two times.

So, arrangements are = 5!/2! 
= 60 ways

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,
E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n(E) = 27
∴P(E) = n(E)/n(S)
= 27/36
= 3/4

Question: A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Solution:
The student can choose 4 questions from first 5 questions or he can also choose 5 questions from the first five questions.

∴ No. of choices available to the student = ⁵C₄ × ⁸C₆ + ⁵C₅ × ⁸C₅
= 5 × 28 + 1 × 56
= 196

Question: If ⁵⁶P_{r + 6} : ⁵⁴P_{r + 3} = 30800 : 1 then the value of r is?

Solution:
⁵⁶Pr + 6 : ⁵⁴P_{r + 3} = 30800 : 1
⇒ 56!/(50 - r)! = (30800 × 54!)/(51 - r!)
⇒ 56 × 55 = 30800/(51 - r)
⇒ 51 - r = 10
∴ r = 41

Question: A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is-

Solution:
Let, S be the sample space

Then,
n(S) = number of ways of drawing 3 balls out of 15
= ¹⁵C₃
= (15 × 14 × 13)/(3 × 2 × 1)
= 455

Let, E = event of getting all the 3 red balls
∴ n(E) = ⁵C₃
= ⁵C₂
= (5 × 4)/(2 × 1)
= 10

∴ P(E) = n(E)/n(S)
= 10/455
= 2/91

Question: In how many ways can 7 persons be seated at a round table if 2 particular persons must not sit next to each other?

Solution:
Total no. of unrestricted arrangements = (7 – 1)! = 6!
When two particular person always sit together, the total no. of arrangements = 6! - 2 × 5!
Required no. of arrangements = 6! - 2 × 5!
= 5! (6 - 2)
= 5 × 4 × 3 × 2 × 4
= 480

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 one head = {HT, TH, HH}

∴ P(E) = n(E)/n(S)
= 3/4

Question: There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is

Solution:
Since there are 4 candidates for the post of a lecturer and 5 men voting, each voter can choose one candidate from the 4. The number of ways in which the votes can be given can be calculated by considering that each of the 5 men can choose one candidate.

For each man, there are 4 possible choices of candidates. Therefore, for 5 men, the total number of ways the votes can be given = 4⁵ = 1024

Question: 8 men entered a lounge simultaneously. If each person shook hands with the other, then find the total no. of hand shakes?

Solution:
Required no. of hand shakes = ⁸C₂
= 8!/2!(8 - 2)!
= 8!/2! · 6!
= 28

Question: When two dice are rolled, what is the probability that the sum of the numbers appeared on them is 11?

Solution:
n(S) = 6²
= 36
n(E) = {(5, 6), (6, 5)} = 2

∴ p(E) = n(E)/n(S)
= 2/36
= 1/18

Question: Using the digits 9, 8, 2, 5 exactly once, how many numbers greater than 5000 can be formed?

Solution:
To form a number greater than 5000, the first digit must be one of 5, 8, or 9.
If the first digit is 5, the remaining 3 digits can be arranged in ³P₃ = 6 ways.
If the first digit is 8, the remaining 3 digits can be arranged in ³P₃ = 6 ways.
If the first digit is 9, the remaining 3 digits can be arranged in ³P₃ = 6 ways.

Therefore, the total number of ways to form such numbers is: 6 + 6 + 6 = 18
18 numbers greater than 5000 can be formed.

Question: In a quality control test, if the probability that a laptop battery lasts 5 years is 5/6, and the probability that its screen remains defect-free for 5 years is 4/5, what is the probability that both the battery and screen will be functioning perfectly after 5 years?

Solution:
Let's
P(Battery) = Probability of battery lasting 5 years = 5/6
P(Screen) = Probability of screen lasting 5 years = 4/5

Required probability = P(Battery) × P(Screen)
= (5/6) × (4/5)
= 20/30
= 2/3

Therefore, there is a 2/3 (or approximately 67%) probability that both the battery and screen will still be functioning perfectly after 5 years.