Discussion
If X is distributed as Binomial \((n, p)\), \(0 \lt p \lt 1\), then :
Answer: B
The true statement is \(n - X\) is distributed \(B(n, 1 - p)\).
X counts the successes in n independent trials with success probability p. Then \(n - X\) counts the failures, and each trial is a failure with probability \(1 - p\), independently. So \(n - X \sim B(n, 1 - p)\). Formally, \(P(n - X = k) = P(X = n - k)\), which is \(\binom{n}{n-k}p^{n-k}(1 - p)^{k} = \binom{n}{k}(1 - p)^{k}p^{n-k}\).
The others do not make sense as binomial laws. \(-X\) takes negative values, and \(B(-n, p)\) has no meaning. \(X - k\) is a shift that can be negative, while \(B(n - k, p)\) lives on \(0, \ldots, n - k\). \(1/X\) is not integer-valued (and is undefined when X = 0), and \(1/p \gt 1\) cannot be a probability.
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