Sum of Two Dice Probabilities with Table (A) - Free Printable

Let’s solve this step by step.

We are rolling two dice. Each die has numbers 1 through 6. When you roll two dice, there are 6 × 6 = 36 total possible outcomes (because each die is independent).

The table in the image shows all possible sums of the two dice — from 2 (1+1) to 12 (6+6). We can count how many ways each sum can happen:

- Sum = 2 → only (1,1) → 1 way
- Sum = 3 → (1,2), (2,1) → 2 ways
- Sum = 4 → (1,3), (2,2), (3,1) → 3 ways
- Sum = 5 → (1,4), (2,3), (3,2), (4,1) → 4 ways
- Sum = 6 → (1,5), (2,4), (3,3), (4,2), (5,1) → 5 ways
- Sum = 7 → (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways
- Sum = 8 → (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways
- Sum = 9 → (3,6), (4,5), (5,4), (6,3) → 4 ways
- Sum = 10 → (4,6), (5,5), (6,4) → 3 ways
- Sum = 11 → (5,6), (6,5) → 2 ways
- Sum = 12 → (6,6) → 1 way

Total: 1+2+3+4+5+6+5+4+3+2+1 = 36 ✔️

Now, probability = (number of favorable outcomes) / 36

Let’s go one by one:

---

P(>2) → sum greater than 2 → that means sum = 3 to 12
Total outcomes for sum > 2 = 36 - (outcomes for sum=2) = 36 - 1 = 35
→ P(>2) = 35/36

P(<11) → sum less than 11 → sum = 2 to 10
Outcomes for sum=11 and 12: 2 + 1 = 3
So outcomes <11 = 36 - 3 = 33
→ P(<11) = 33/36 = 11/12

P(≥2) → sum ≥ 2 → ALL outcomes (since minimum sum is 2)
→ P(≥2) = 36/36 = 1

P(≥12) → sum ≥ 12 → only sum=12 → 1 outcome
→ P(≥12) = 1/36

P(≤7) → sum ≤ 7 → sums 2 through 7
Count: 1+2+3+4+5+6 = 21
→ P(≤7) = 21/36 = 7/12

P(<5) → sum < 5 → sums 2,3,4 → 1+2+3 = 6
→ P(<5) = 6/36 = 1/6

P(≤5) → sum ≤ 5 → sums 2,3,4,5 → 1+2+3+4 = 10
→ P(≤5) = 10/36 = 5/18

P(10) → exactly sum=10 → 3 outcomes
→ P(10) = 3/36 = 1/12

P(≥8) → sum ≥ 8 → sums 8,9,10,11,12 → 5+4+3+2+1 = 15
→ P(≥8) = 15/36 = 5/12

P(<6) → sum < 6 → sums 2,3,4,5 → same as P(≤5) above? Wait no: <6 means up to 5 → yes, same as P(≤5) → 10 outcomes
Wait! Actually, <6 means 2,3,4,5 → yes, 10 outcomes → 10/36 = 5/18
But let me double-check: sum=2,3,4,5 → 1+2+3+4=10 → correct.

Actually wait — I think I made a mistake here. Let me recheck:

P(<6) → sum strictly less than 6 → so 2,3,4,5 → yes, 10 outcomes → 10/36 = 5/18

But earlier I did P(≤5) also as 10/36 — that’s correct because ≤5 includes 5, and <6 also includes 5 — so they are the same? No!

Wait: ≤5 means 2,3,4,5 → 10 outcomes
<6 means 2,3,4,5 → same thing! Because next integer after 5 is 6, so <6 is same as ≤5 when dealing with integers.

Yes! So P(<6) = P(≤5) = 10/36 = 5/18

But let’s keep going.

P(≥12) → already done → 1/36

P(≤10) → sum ≤ 10 → exclude 11 and 12 → 36 - (2+1) = 33
→ P(≤10) = 33/36 = 11/12

P(≥6) → sum ≥ 6 → sums 6 to 12
Count: 5 (for 6) + 6 (7) + 5 (8) + 4 (9) + 3 (10) + 2 (11) + 1 (12) = 5+6=11; 11+5=16; 16+4=20; 20+3=23; 23+2=25; 25+1=26
→ 26 outcomes
→ P(≥6) = 26/36 = 13/18

P(≤12) → all outcomes → 36/36 = 1

P(≥11) → sum ≥ 11 → 11 and 12 → 2 + 1 = 3
→ P(≥11) = 3/36 = 1/12

P(>5) → sum > 5 → sums 6 to 12 → same as P(≥6) above?
>5 means 6,7,8,...,12 → yes, same as ≥6 → 26 outcomes → 26/36 = 13/18

Wait — let me confirm:

Sum > 5 → starts at 6 → yes, same as ≥6 → 26 outcomes → correct.

Now let’s list all answers clearly:

Left column:
- P(>2) = 35/36
- P(≥2) = 1
- P(≤7) = 21/36 = 7/12
- P(≤5) = 10/36 = 5/18
- P(≥8) = 15/36 = 5/12
- P(≥12) = 1/36
- P(≥6) = 26/36 = 13/18
- P(≥11) = 3/36 = 1/12

Right column:
- P(<11) = 33/36 = 11/12
- P(≥12) = 1/36
- P(<5) = 6/36 = 1/6
- P(10) = 3/36 = 1/12
- P(<6) = 10/36 = 5/18
- P(≤10) = 33/36 = 11/12
- P(≤12) = 1
- P(>5) = 26/36 = 13/18

I notice P(≥12) appears twice — both times it's 1/36 — that’s fine.

Also, P(≤12) = 1, which makes sense.

Let me verify a few more:

P(≤7): sums 2 to 7 → 1+2+3+4+5+6 = 21 → 21/36 = 7/12 ✔️

P(≥8): 8 to 12 → 5+4+3+2+1=15 → 15/36=5/12 ✔️

P(>5): 6 to 12 → 5+6+5+4+3+2+1=26 → 26/36=13/18 ✔️

All look good.

Final Answer:

P(>2) = 35/36
P(≥2) = 1
P(≤7) = 7/12
P(≤5) = 5/18
P(≥8) = 5/12
P(≥12) = 1/36
P(≥6) = 13/18
P(≥11) = 1/12

P(<11) = 11/12
P(≥12) = 1/36
P(<5) = 1/6
P(10) = 1/12
P(<6) = 5/18
P(≤10) = 11/12
P(≤12) = 1
P(>5) = 13/18