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Substitute the values:
P(2) = \binom{4}{2} (0.3)^2 (0.7)^2 = 6 \cdot (0.09) \cdot (0.49) = 6 \cdot 0.0441 = 0.2646
Thus, the probability that exactly two of the four rolls result in a number greater than 7 is:

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