If one rolls a fair sides die twice, what is the probability that the die will land on the same number on both the occasions?

The five letters could be arrange in 5! ways.
One of them is 'BRING'.
? Required probability = 1/5!
= 1/(5 x 4 x 3 x 2 x 1)
= 1/120

Let the name of the man be A and that of his wife be B. Then ,P(A will not be alive after 10 yr and B will not be alive after 10yr) = 3/4 x 2/3 = 1/2

Time taken for complete cycle = 60 s
Probability that light will be green = (Time for which light is green)/(time taken for complete cycle)
= 25/60 = 5/12
? Probability that light will not be green = 1- 5/12 = 7/12

Total number of ways of selecting one number from 25 number is ²⁵C₁ ways.
Let E be the event of drawing a number divisible by both 2 and 3
i.e., (if a number is divisible by both 2 and 3, it is divisible by 6 also)
E = {6, 12, 18, 24}
? Required probability = P(E) = n(E) / n(S) = 4/25

Here, n(S) = 6 x 6 = 36 and E = Event of getting a number other than 4 on any dice
= {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2) , (2, 3) , (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (5, 1) , (5, 2) , (5, 3), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 5), (6, 6) }
? P(E) = n(E) / n(S) = 25/36

Let N be the number of coins showing heads.
? Probability when N equal to 50 = Probability when N equal to 51
? ¹⁰⁰C₅₀ x P ⁵⁰ x (1 - p)⁵⁰ = ¹⁰⁰C₅₁ x (1 - p)⁴⁹
? p = 51/101

If the product of the four numbers ends in one of the digits 1, 3, 7 or 9, each number should have the last digit as one of these 4 digits.
? The number of favourable cases = 4⁴
Total number of all possible cases = 10⁴
Hence, the required probability =4⁴/10⁴
= 2⁴/5⁴ = 16/625

Required probability = 3/52 + 13/52 = 16/52 =4/13
[Hint Why 13/52 because there are 13 spades and why 3/52 instead of 4/52 (there are four kings) because one king is already counted in spades.]

Total ways = 52
There is one queen of club and one king of heart.
? Favorable ways = 1 + 1 = 2
? Required probability = 2/52 = 1/26

n(S) = 2³ = 8
Let E = Event of getting exactly two heads
= {(H, H, T), (H, T ,H), (T, H, H)}
? n(E) = 3
? required probability = 3/8