Number 1 to 25 are marked on tokens of equals size one on each. A token is drawn at random. Find the probability of getting a number divisible by both 2 and 3?

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