Let E = Event of getting same number on both the occasions
= (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Correct Option: B

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

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

Correct Option: D

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

Correct Option: A

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

Correct Option: B

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

Correct Option: B

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