If 3 dice are thrown, what is the probability that at least one dice shows 1 or 2? radio_button_unchecked 19/27 radio_button_unchecked 64/216 radio_button_unchecked 1/9 radio_button_unchecked 1/3

To solve this problem, it's easier to calculate the probability of the opposite event (none of the dice shows 1 or 2) and then subtract it from 1.

Step 1: Calculate the probability of not getting a 1 or 2 on a single die. There are 4 favorable outcomes (3, 4, 5, 6) out of 6 possible outcomes. So, the probability is 4/6 = 2/3.

Step 2: Since the dice are independent, the probability of not getting a 1 or 2 on three dice is (2/3) * (2/3) * (2/3) = 8/27.

Step 3: The probability of getting at least one 1 or 2 is the complement of the event calculated in step 2. So, the probability is 1 - 8/27 = 19/27.

So, the correct answer is:

radio_button_checked 19/27

radio_button_unchecked 64/216

radio_button_unchecked 1/9

radio_button_unchecked 1/3