Two numbers are selected at random from $1,2,3,...,100$ and multiplied.  Then the probability that the product thus obtained is divisible by 3 is

The Random experimnet is " Selecting two numbers at random from the set of first 100 natural numbers" 

Let $S$ be the sample space, $n\left(S\right)={100}_{C_2}$

Let $E$ be the event of getting numbers whose product is divisible by 3. 

Hence, the event $\overline{E}$ be complement of event, getting two numbers which are not divisible by 3

hence, the number of favorable cases to the event $\overline{E}$ is $n\left(\overline{E}\right)={67}_{C_2}$

Hence, 

$\begin{array}{rcl}P\left(\overline{E}\right)& =& \frac{n\left(\overline{E}\right)}{n\left(S\right)}\\ & =& \frac{{67}_{C_2}}{{100}_{C_2}}\\ & =& \frac{67·66}{100·99}\\ & =& \frac{67}{150}\end{array}$

Therefore, the required probability is 

    $\begin{array}{rcl}P\left(E\right)& =& 1-P\left(\overline{E}\right)\\ & =& 1-\frac{67}{150}\\ & =& \boxed{\frac{83}{150}}\end{array}$