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You can use a majority logic; for a single info bit, find whether the code bits are mostly ones or zeros? If ones are more your decision is 1 and vice versa.
e.g. 1 (info bit) -> 1 1 1 (1/3 repetition code) -> 0.8 0.9 -0.1 (received bits) -> 1 1 0 (decoded) -> 1 (decision bec. we have two 1's and one 0)