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Since two cards are drawn with replacement from 6 cards (numbered 1 to 6), the total number of outcomes is $6 \times 6 = 36$.
Event A is the sum of numbers being 10. The possible pairs are (4, 6), (5, 5), and (6, 4). Thus, $P(A) = \frac{3}{36} = \frac{1}{12}$.
Event B is getting a number other than 4 on the first card. The first card can be 1, 2, 3, 5, or 6 (5 options). The second card can be any of the 6 numbers. Thus, $n(B) = 5 \times 6 = 30$. $P(B) = \frac{30}{36} = \frac{5}{6}$.
Event (A and B) means the sum is 10 AND the first card is not 4. From the pairs in A {(4, 6), (5, 5), (6, 4)}, only (5, 5) and (6, 4) satisfy the condition of the first card not being 4. Thus, $P(A \cap B) = \frac{2}{36} = \frac{1}{18}$.
Events are independent if $P(A \cap B) = P(A) \times P(B)$. Here, $P(A) \times P(B) = \frac{1}{12} \times \frac{5}{6} = \frac{5}{72}$. Since $\frac{1}{18} \neq \frac{5}{72}$, the events are not independent.
Final Answer: P(A and B) = 1/18; Events are not independent.
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