The five letters could be arrange in 5! ways.
One of them is 'BRING'.

Correct Option: A

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

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.]

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

Correct Option: C

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

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

Total Number of outcomes = ¹⁰C₃ = 120
Number of outcomes not containing red balls = ⁵C₃ = 10
∴ Probability that at least one is red = 1 - 10/120

Correct Option: C

Total Number of outcomes = ¹⁰C₃ = 120
Number of outcomes not containing red balls = ⁵C₃ = 10
∴ Probability that at least one is red = 1 - 10/120 = 11/12