If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of then are divisible by 2 or 3, is

If three distinct numbers are chosen randomly from the first 100 natural numbers , then the probability that all three of them are divisible by 2 and 3 is :

If three distinct number are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is a. 4//25 b. 4//35 c. 4//33 d. 4//1155

If three distinct number are chosen randomly from the first 100 natural numbers,then the probability that all three of them are divisible by both 2 and 3 is (a) (4)/(25) (b) (4)/(35) (c) (4)/(33) (d) (4)/(1155)

If three distinct number are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is (a) 4/25 (b) 4/35 (c) 4/33 (d) 4/1155

If three distinct number are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is (a) 4/25 (b) 4/35 (c) 4/33 (d) 4/1155

If three distinct number are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is (a) 4/25 (b) 4/35 (c) 4/33 (d) 4/1155

If 4 distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all 4 of them are either divisible by 3 or divisible by 5 is

If 4 distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all 4 of them are either divisible by 3 or divisible by 5 is