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(A) Let $D_1$ be the first die and $D_2$ be the second die. The outcomes for $D_1$ are ${1, 1, 2, 2, 3, 4}$ and for $D_2$ are ${1, 2, 2, 3, 3, 4}$. To

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$\frac{1}{2}$

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(A) Let $D_1$ be the first die and $D_2$ be the second die. The outcomes for $D_1$ are ${1, 1, 2, 2, 3, 4}$ and for $D_2$ are ${1, 2, 2, 3, 3, 4}$. Total outcomes $= 6 	imes 6 = 36$.We want the sum $S = 4$ or $S = 5$.For $S = 4$,the possible pairs $(D_1, D_2)$ are $(1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (4, 	ext{none})$.Counting frequencies: $(1, 3)$ occurs $2 	imes 2 = 4$ times,$(2, 2)$ occurs $2 	imes 2 = 4$ times,$(3, 1)$ occurs $1 	imes 1 = 1$ time. Total for $S=4$ is $4+4+1 = 9$.For $S = 5$,the possible pairs $(D_1, D_2)$ are $(1, 4), (2, 3), (2, 3), (3, 2), (3, 2), (4, 1)$.Counting frequencies: $(1, 4)$ occurs $2 	imes 1 = 2$ times,$(2, 3)$ occurs $2 	imes 2 = 4$ times,$(3, 2)$ occurs $1 	imes 2 = 2$ times,$(4, 1)$ occurs $1 	imes 1 = 1$ time. Total for $S=5$ is $2+4+2+1 = 9$.Total favorable outcomes $= 9 + 9 = 18$.Probability $= \frac{18}{36} = \frac{1}{2}$.

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