Write the face distributions of the two dice

Let die $A$ have faces: $1,1,2,2,3,4$ So, $P_A(1)=\frac{2}{6}=\frac{1}{3},\quad P_A(2)=\frac{2}{6}=\frac{1}{3},\quad P_A(3)=\frac{1}{6},\quad P_A(4)=\frac{1}{6}.$

Let die $B$ have faces: $1,2,2,3,3,4$ So, $P_B(1)=\frac{1}{6},\quad P_B(2)=\frac{2}{6}=\frac{1}{3},\quad P_B(3)=\frac{2}{6}=\frac{1}{3},\quad P_B(4)=\frac{1}{6}.$

Total number of equally likely outcomes

Each die has $6$ faces, so total outcomes when thrown together: $6\times 6=36.$

Find outcomes giving sum $4$

Sum $4$ can occur as:

• $(1,3)$
• $(2,2)$
• $(3,1)$

Count each using repeated faces:

• For $(1,3)$: die $A$ has two $1$'s and die $B$ has two $3$'s $2\times 2=4$
• For $(2,2)$: die $A$ has two $2$'s and die $B$ has two $2$'s $2\times 2=4$
• For $(3,1)$: die $A$ has one $3$ and die $B$ has one $1$ $1\times 1=1$

Hence total ways for sum $4$: $4+4+1=9.$

Find outcomes giving sum $5$

Sum $5$ can occur as:

• $(1,4)$
• $(2,3)$
• $(3,2)$
• $(4,1)$

Count each:

• For $(1,4)$: die $A$ has two $1$'s and die $B$ has one $4$ $2\times 1=2$
• For $(2,3)$: die $A$ has two $2$'s and die $B$ has two $3$'s $2\times 2=4$
• For $(3,2)$: die $A$ has one $3$ and die $B$ has two $2$'s $1\times 2=2$
• For $(4,1)$: die $A$ has one $4$ and die $B$ has one $1$ $1\times 1=1$

Hence total ways for sum $5$: $2+4+2+1=9.$

Total favorable outcomes

$9+9=18.$

Probability

$P(\text{sum is }4\text{ or }5)=\frac{18}{36}=\frac{1}{2}.$

Check options

The correct option is: $\boxed{\text{D }\frac{1}{2}}$

Comparison with stored answer

Stored correct answer is D, which matches our result.