In order to have independent events, we need to verify the following formula:

For event A, the outcomes that are less than 6 are 1, 2, 3, 4 and 5, so the probability of event A is:

For event B, the outcomes that are divisible by 3 are 3 and 6, so:

The intersection of these events is only number 3, so:

Verifying the formula, we have:

The result is false, so the events are not independent, therefore the answer is NO.