n(S) = 6 x 6 = 36
n(E) = (1, 2) (2, 1), (1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (3, 6) (4, 5) (5, 4) (6, 3) (6, 6) = 12

Correct Option: B

n(S) = 6 x 6 = 36
n(E) = (1, 2) (2, 1), (1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (3, 6) (4, 5) (5, 4) (6, 3) (6, 6) = 12
∴ P(E) = 12/36

Required probability = Probability of getting 2 tails + Probability of getting 1 tail + Probability of getting no tail.

Correct Option: A

Required probability = Probability of getting 2 tails + Probability of getting 1 tail + Probability of getting no tail.
= ⁸C₂ x 1/256 + ⁸C₁ x 1/256 + ⁸C₀ x 1/256
= 37/256

Required probability = P(A). P(B) + P (A ) . P(B)

Correct Option: B

Required probability = P(A). P(B) + P (A ) . P(B)
= 5/7 . 3/10 + 2/7 . 7/10
= 29/70

Probability that first ball drawn is white = ⁵C₁ = 1/4
Since, balls are drawn with replacement, hence all the four events will have equal probability.

Correct Option: A

Probability that first ball drawn is white = ⁵C₁ = 1/4
Since, balls are drawn with replacement, hence all the four events will have equal probability.
Therefore, required probability = 1/4 x 1/4 x 1/4 x 1/4 = 1/256

Total number of balls in the bag = 12
Probability of drawing one blue ball in the first draw P(E₁) = ⁸C₁ / ¹²C₁
= 8/12 = 2/3

After the first drawn of a blue ball, now there are 7 blue and 4 white ball in the bag. Total number of the balls in the bag is 11.

Probability of drawing one blue ball in the second drawn = P(E₂) = ⁷C₁ /¹¹C₁ = 7/11

Correct Option: B

Total number of balls in the bag = 12
Probability of drawing one blue ball in the first draw P(E₁) = ⁸C₁ / ¹²C₁
= 8/12 = 2/3

After the first drawn of a blue ball, now there are 7 blue and 4 white ball in the bag. Total number of the balls in the bag is 11.

Probability of drawing one blue ball in the second drawn = P(E₂) = ⁷C₁ /¹¹C₁ = 7/11

∴ Probability that both are blue P(E₁ ∩ E₂) = P(E₁) x P(E₂)
= 2/3 x 7/11 = 14 / 33