## Answer :

We'll start by considering all possible outcomes when rolling two dice. Each die has 6 faces, so when rolling two dice, there are a total of [tex]\(6 \times 6 = 36\)[/tex] possible outcomes.

### (a) Calculating Probability [tex]\(P(A)\)[/tex]

Event [tex]\(A\)[/tex] is the event where the sum of the two dice is greater than 9. We need to determine the successful outcomes for this event.

The possible sums greater than 9 are: [tex]\(10, 11, \)[/tex] and [tex]\(12\)[/tex].

Let's list the combinations for each sum:

- Sum of 10: (4,6), (5,5), (6,4) → 3 outcomes

- Sum of 11: (5,6), (6,5) → 2 outcomes

- Sum of 12: (6,6) → 1 outcome

Adding these together, we get:

[tex]\[ 3 + 2 + 1 = 6 \text{ successful outcomes} \][/tex]

To find the probability, we divide the number of successful outcomes by the total number of outcomes:

[tex]\[ P(A) = \frac{6}{36} = \frac{1}{6} \approx 0.17 \][/tex]

So,

[tex]\[ P(A) \approx 0.17 \][/tex]

### (b) Calculating Probability [tex]\(P(B)\)[/tex]

Event [tex]\(B\)[/tex] is the event where the sum is neither divisible by 4 nor by 5.

First, let's find the totals that are either divisible by 4 or by 5.

- Sums divisible by 4: [tex]\(4, 8, 12\)[/tex]

- Combinations for 4: (1,3), (2,2), (3,1) → 3 outcomes

- Combinations for 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes

- Combinations for 12: (6,6) → 1 outcome

Adding these together, we get:

[tex]\[ 3 + 5 + 1 = 9 \text{ outcomes where sum is divisible by 4} \][/tex]

- Sums divisible by 5: [tex]\(5, 10\)[/tex]

- Combinations for 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes

- Combinations for 10: (4,6), (5,5), (6,4) → 3 outcomes

Adding these together, we get:

[tex]\[ 4 + 3 = 7 \text{ outcomes where sum is divisible by 5} \][/tex]

Now, let's count the overlap (sums divisible by both 4 and 5, i.e., [tex]\(20\)[/tex]), but as neither 4 nor 5 can appear as results of two dice roll, there are no overlaps seemingly.

However, note that sum [tex]\(20\)[/tex] can't be twice of sum of a 6-sided die.

Thus, we have identified:

- Total outcomes divisible by 4: 9

- Total outcomes divisible by 5: 7

Since these sets are mutually exclusive (since their overlap is 0):

[tex]\[ \text {Total outcomes divisible by 4 or 5} = 9 + 7 = 16 \][/tex]

To find sums

**not**divisible by 4 or 5, we subtract these from the total outcomes:

[tex]\[ \text{Total outcomes not divisible by 4 or 5} = 36 - 16 = 20 \][/tex]

Finally, the probability for Event [tex]\(B\)[/tex] is:

[tex]\[ P(B) = \frac{20}{36} \approx 0.56 \][/tex]

Therefore, rounding to two decimal places:

[tex]\[ (a) \, P(A) \approx 0.17 \][/tex]

[tex]\[ (b) \, P(B) \approx 0.56 \][/tex]