Question

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

Answer 1

no. div by 2 and 3 is div by 6.
we have 16 such no.s {6,12,18,.......,96}
$\frac{16c3}{100c3}$
c= C of nCr

Answer 2

The probability that all three of them are divisible by 2 and 3 is $\frac{4}{1155}$.

Step-by-step explanation:

Solution:

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 }

n = $\frac{96}{6}$

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 favorable choices = $16 \times 15 \times 14$ = n(A)

Total no.of ways= $100 \times 99 \times 98$=n(S)

P(divisible by 2 and 3)= $\frac{n(A)}{n(S)}$

= $\frac{( 100 \times 99 \times 98)}{( 16 \times 15 \times 14)}$

= $\frac{4}{1155}$.