Discussion
If X is \(F(m, n)\) (F-distribution with m and n degrees of freedom) and Y is \(F(n, m)\), given that \(P[X \ge c] = a\) then \(P\left[Y \ge \dfrac{1}{c}\right]\) is :
Answer: D
\(P[Y \ge 1/c] = \)\(1 - a\).
If \(X \sim F(m, n)\), then \(1/X \sim F(n, m)\): the F variable is a ratio of two scaled chi-squares, and inverting it swaps numerator and denominator. So Y has the same distribution as \(1/X\), and for \(c \gt 0\)
\[ \begin{aligned} P\left[Y \ge \frac{1}{c}\right] &= P\left[\frac{1}{X} \ge \frac{1}{c}\right] \\ &= P[X \le c] \\ &= 1 - P[X \ge c] = 1 - a \end{aligned} \]
(the F distribution is continuous, so \(P[X = c] = 0\)).
This is the identity behind printed F tables: the lower percentage point of \(F(m, n)\) is the reciprocal of the upper point of \(F(n, m)\). Answer a forgets that inverting reverses the inequality, and 1 or \(a/2\) have no basis.
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