Total balls in the box = 5
Second red ball can be drawn in two ways
Case I. First ball is white and second ball is red.
Its probability = 3/5.2/4 = 6/20 = 3/10
Case II: First ball is red and second ball is red
Its probability = 2/5.1/4 = 2/20 = 1/10

Correct Option: B

Total balls in the box = 5
Second red ball can be drawn in two ways
Case I. First ball is white and second ball is red.
Its probability = 3/5.2/4 = 6/20 = 3/10
Case II: First ball is red and second ball is red
Its probability = 2/5.1/4 = 2/20 = 1/10
Hence, reqd. probability
= 3/10 + 1/10 = 4/10 = 2/5

Total number of ways
= n(s) = ⁶c₂ = 15
Favorable number of ways
= n(E) = ³C₁ x ³C₁= 9

Correct Option: B

Total number of ways
= n(s) = ⁶c₂ = 15
Favorable number of ways
= n(E) = ³C₁ x ³C₁= 9
∴ Required probability = 9/15 = 3/5

Total ways = ¹³c₂
Favourable number ways of selecting men only = ⁸c₂
∴ Probability of selecting no woman
= ⁸c₂ / ¹³c₂
= 14/39

Correct Option: A

Total ways = ¹³c₂
Favourable number ways of selecting men only = ⁸c₂
∴ Probability of selecting no woman
= ⁸c₂ / ¹³c₂
= 14/39
∴ Probability of selecting at least one woman
= 1 - (14 / 39)
= 25 / 39

Let P(B) = x
Given, P(A∪B) = 0.8 and P(A) = 0.3
⇒ P(A) + P(B) - P(A∩B) = 0.8
⇒ P(A) + P(B) - P(A) P(B) = 0.8 {∵A and B are independent}

Correct Option: E

Let P(B) = x
Given, P(A∪B) = 0.8 and P(A) = 0.3
⇒ P(A) + P(B) - P(A∩B) = 0.8
⇒ P(A) + P(B) - P(A) P(B) = 0.8 {∵A and B are independent}
⇒ 0.3 + x - 0.3x = 0.8
⇒ 0.7x = 0.5
∴ x = 5/7

Total events = n (s) = 2⁵ = 32
n(E) of getting heads = 1
p(E) = 1/32
∴ n(E) = 1 - p(E)

Correct Option: A

Total events = n (s) = 2⁵ = 32
n(E) of getting heads = 1
p(E) = 1/32
∴ n(E) = 1 - p(E) = 1 - 1/32 = 31/32