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

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

Grade:Upto college level

2 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
Please find the answer to your question below
A natural number is divisible by 2 or 3, if it is divisible by 6.
So, numbers divisible by 6 are 6,12,18, . . . . ,96
So there are total 16 numbers out of which 3 distinct number can be chosen in 16×15×14=n(A)
Total no of ways=100×99×98=n(S)
Required probability=n(A)/n(S)=4/1155 .
Kaushik Sarkar
15 Points
5 years ago
Any natural number is divisible by both 2 and 3 if it is divisible by (2×3=) 6
 
So, the numbers divisible by 6 from 1 to 100 are 6,12,18,....96
96=6+(n-1)6  {where 'n' is no.of terms }
On solving, we get n=16 
 
So, there are total 16 numbers out of which the 3 
distinct numbers are to be chosen.
 
Total no. of favourable choices =16×15×14=n(A)
Total no.of ways=100×99×98=n(S)
 
P(divisible by 2 and 3)=n(A)/n(S)=(100×99×98)/(16×15×14)=4/1155  (answer)
 
 

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