1. **Problem statement:** We have a deck with two cards each of the integers from -10 to 10, excluding zero. We want the probability of dealing two cards whose product is 12. 2. **Total cards:** The integers are from -10 to -1 and 1 to 10, so 20 distinct integers, each with 2 cards, total cards = $20 \times 2 = 40$. 3. **Total ways to choose 2 cards:** $$\binom{40}{2} = \frac{40 \times 39}{2} = 780$$ 4. **Find pairs with product 12:** Possible integer pairs $(a,b)$ with $a \times b = 12$ are $(1,12)$, $(2,6)$, $(3,4)$ and their negatives. But only integers from -10 to 10 are in the deck, so valid pairs are: - $(2,6)$ and $(6,2)$ - $(3,4)$ and $(4,3)$ - $(-2,-6)$ and $(-6,-2)$ - $(-3,-4)$ and $(-4,-3)$ Note: 12 and -12 are not in the deck. 5. **Count cards for each number:** Each number has 2 cards. 6. **Count favorable pairs:** For each pair of distinct numbers, number of ways to pick one card from each is $2 \times 2 = 4$. Pairs: - $(2,6)$: 4 ways - $(3,4)$: 4 ways - $(-2,-6)$: 4 ways - $(-3,-4)$: 4 ways Total favorable ways = $4 + 4 + 4 + 4 = 16$ 7. **Calculate probability:** $$P = \frac{16}{780}$$ 8. **Reduce fraction:** Divide numerator and denominator by 4: $$\frac{16}{780} = \frac{\cancel{4} \times 4}{\cancel{4} \times 195} = \frac{4}{195}$$ 9. **Decimal:** $$\frac{4}{195} \approx 0.02051$$ 10. **Percent:** $$0.02051 \times 100 = 2.051\%$$ **Final answer:** Probability = $\frac{4}{195}$, approximately 0.02051, or 2.051%.