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## Probability calculations

The 12 listed stars are part of the 163 objects the
magnitude of which is less than 3. The average error,
on both sides of the ideal position,
is worth 0. 263d. Each star corresponds therefore to one
strip of celestial sphere the extension
of which equals 0. 526d.

The 163 stars represent 85. 738d.

These 85. 738d represent 47. 6% of the celestial sphere
(85. 738 / 180).

The probability there is to observe the zenith culmination of one of
those stars in a given place (with a mean error of 0. 263d)
equals therefore 0. 476.

If we take into account 12 stars, the probability equals 0. 476
^{12} = **0. 0001** .

If we take into account the four brightest stars of the list
(Arcturus, Spica, Antares and Fomalhaut), the probability
equals **1. 21 / 10**^{5}.

Mean error = 0. 295d

The 18 stars the magnitude of which is less than 1. 25 represent
5. 9 % of the celestial sphere: (2 * 0. 295 * 18) / 180 = 0. 059

0. 059^{4} = 1. 21 / 10^{5}

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