Period of cot 3x is π/3 and period of cos (4x + 3) is π/2

⇒ L.C.M. is π, hence the answer.

sin 30° + cos 60° + tan 45°

= (1/2) + (1/2) + 1

= 2

Since, sin 190° = - sin 10°, sin 200° = - sin 20°,

sin 210° = - sin 30°, sin 360° = sin 180° = 0 etc.

Hence all the terms in the expression cancels out, therefore answer is 0.

Let ABCDEF be a regular hexagon.

Three vertices out of its 6 vertices can be chosen in ⁶C₃ ways. Therefore, 20 triangles can be made by joining three vertices at a time.

Out of these 20 triangles, only ACE and BDF are equilateral. Hence, the required probability = 1/10.

Probability of getting atleast seven points

= Probability of getting 7 points or 8 points

= Prob. of getting 7 points + Prob. of getting 8 points.

Seven points in four matches can be obtained in the following four different ways :

2, 2, 2, 1;

2, 2, 1, 2;

2, 1, 2, 2;

1, 2, 2, 2

The probability of each of these ways = (0.50)^{3 x} (0.05) (by multiplication theorem for independent events)

= 0.00625

Hence, probability of getting 7 points = ^{4 x} (0.00625) (by add. Theorem)

= 0.0250.

Eight points in four matches can be obtained only in one way i.e. 2, 2, 2, 2.

Hence, Prob. of getting 8 points = (0.50)⁴ = 0.0625

Thus, the required prob. = 0.250 + 0.0625 = 0.0875.

Probability that a man will live 10 more years = 1/4

Probability that man not live 10 more years = 1 - 1/4 = 3/4

Probability that his wife will live 10 more years = 1/3

Probability that his wife not live 10 more years = 1 - 1/3 = 2/3

Then, probability that neither will be alive in 10 years = (3/4) x (2/3) = 1/2

Exhaustive number of cases = 6³ = 216.

The same number can appear on each of the dice in the following ways: (1, 1, 1), (2, 2, 2).... (6, 6, 6).

So, favourable number of cases = 6.

Hence, required probability = 6/216 = 1/36.