What is the probability that a card drawn at random from a pack of 52 cards is either a king or a spade?

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

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

Let E = Event of getting same number on both the occasions
= (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
? n(E) = 6
? Required probability = 6/6² = 1/6

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

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

n(S) = 49
Favourable numbers are 11, 21, 31, 41.
? Required probability = 4/49

n(s) = 50
Prime numbers are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
? n(E) = 15
? p(E) = 15/50 = 3/10 = 0.3

Required probability = (²C₁ x ³C₂ + ²C₂ x ³C₁) / (⁵C₃) = 9/10