 Solution (source code)

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= Solution

No. The high-dose expectation is
$$
E_B=\frac12+\frac12+3\frac45=3.4.
$$
Although both groups start with five patients, censoring removes two low-dose patients at month 3 and one high-dose patient at month 1. Their later risk sets therefore differ; only the sum $E_A+E_B=5$ must equal the total observed failures.

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