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There are two containers, with one containing 4 red and 3 green balls and the other containing 3 blue and 4 green balls. One ball is drawn at random from each container. The probability that one of the balls is red and the other is blue will be

A. $$\frac{1}{7}$$

B. $$\frac{9}{{49}}$$

C. $$\frac{{12}}{{49}}$$

D. $$\frac{3}{7}$$

Answer: Option C

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