A complete cycle of a traffic light takes 60 s. During each cycle the light is green for 25 s, yellow for 5 s and red for 30 s. At a randomly chosen time, the probability that the light will not be green is?

Total number of ways of selecting one number from 25 number is ²⁵C₁ ways.
Let E be the event of drawing a number divisible by both 2 and 3
i.e., (if a number is divisible by both 2 and 3, it is divisible by 6 also)
E = {6, 12, 18, 24}
? Required probability = P(E) = n(E) / n(S) = 4/25

Here, n(S) = 6 x 6 = 36 and E = Event of getting a number other than 4 on any dice
= {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2) , (2, 3) , (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (5, 1) , (5, 2) , (5, 3), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 5), (6, 6) }
? P(E) = n(E) / n(S) = 25/36

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

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

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

Let the name of the man be A and that of his wife be B. Then ,P(A will not be alive after 10 yr and B will not be alive after 10yr) = 3/4 x 2/3 = 1/2

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

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
? 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
Hence, the required probability =4⁴/10⁴
= 2⁴/5⁴ = 16/625