Answer: [tex]\dfrac{1}{3}[/tex]
Step-by-step explanation:
Favorable outcomes (sum is a multiple of 3): (1,2) , (1, 5), (1,8), (2,1) , (2,4), (2,7) , (3, 3), (3, 6), (4, 2), (4, 5) , (4, 8), (5,1), (5, 4), (5, 7), (6,3), (6, 6).
Number of favorable outcomes (sum is a multiple of 3): 16
Total outcomes = Total outcomes of 6-sided die x Total outcomes of 8-sided die=6 × 8 = 48
Then , the probability of getting numbers whose sum is a multiple of 3:
[tex]=\dfrac{\text{favorable outcomes}}{\text{total outcomes}}\\\\=\dfrac{16}{48}=\dfrac{1}{3}[/tex]
Hence, the required probability =[tex]\dfrac{1}{3}[/tex]
