Here, we have S = {1, 2, 3,....., 15}
⇒    n(s) = 15
Let A be the favourable av + causes of getting a multiple 4, then A = {4,8,12}
⇒     n (A) = 3

Therefore, P(A) = 

When a die is thrown once, then possible outcome(s) are 6
i.e.,    n(S) = 6
Let ‘A’ be the favourable outcomes of getting a number greater than 5. Then A = (6) i.e.,    n(A) = 1
Therefore,

Total number of possible outcome(s) are 16 i.e.,    n(S) = 16

Let A be the favourable outcomes of getting getting a number which is perfect square. Then A = (9,16) i.e.,    n(A) = 2

Therefore,

S = {H, T}

When a die is thrown oner then possible outcome(s) are 6
i.e.,    n(S) = 6
Let A be the favourable outcomes of setting a number less than 3. then
A = (1,2) i.e.,    n(A) = 2
Therefore,