One hundred identical coins each with probability p of showing up head are tossed. If 0 < p < 1 and the probability of heads showing on 50 coins is equals to that of heads on 51 coins. Then the value of p, is?

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

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 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

Time taken for complete cycle = 60 s
Probability that light will be green = (Time for which light is green)/(time taken for complete cycle)
= 25/60 = 5/12
? Probability that light will not be green = 1- 5/12 = 7/12

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

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

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.]

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

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

n(S) = 49
Favourable numbers are 11, 21, 31, 41.
? Required probability = 4/49