The problem in the image asks to find the probability that the sum of rolling two dice will be less than 3 or greater than 6.

Step 1: Understand the total number of possible outcomes

When rolling two dice, each die has 6 faces, so the total number of possible outcomes is:

$6 \times 6 = 36$

Step 2: Identify the favorable outcomes

• Sum less than 3: The smallest possible sum is 2, which can only be achieved by rolling a 1 on both dice (1,1). So, there is only 1 favorable outcome for the sum to be less than 3.

• Sum greater than 6: We need to find all the outcomes where the sum is greater than 6. These sums can be:

  • 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
  • 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
  • 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
  • 10: (4,6), (5,5), (6,4) → 3 outcomes
  • 11: (5,6), (6,5) → 2 outcomes
  • 12: (6,6) → 1 outcome

So, the number of favorable outcomes for a sum greater than 6 is:

$6 + 5 + 4 + 3 + 2 + 1 = 21$

Step 3: Add the favorable outcomes

The favorable outcomes for the event "less than 3 or greater than 6" are: $1 \text{ (for less than 3)} + 21 \text{ (for greater than 6)} = 22$

Step 4: Calculate the probability

The probability is the ratio of favorable outcomes to total outcomes:

$\frac{22}{36} = \frac{11}{18}$

Final Answer:

The probability that the sum will be less than 3 or greater than 6 is:

$\frac{11}{18}$

Would you like further details or have any other questions?

Here are some related questions:

1. What is the probability that the sum of the dice will be exactly 7?
2. How many outcomes result in a sum less than or equal to 4?
3. What is the probability of rolling a sum of 9?
4. How would the probability change if the dice were biased?
5. What is the probability of rolling a sum between 4 and 10?

Tip: In problems involving dice, always list all possible outcomes for the sums you are interested in to avoid missing any possibilities!