Write the event before choosing a distribution

Distinguish exact counts, inclusive boundaries, replacement assumptions and observation-window units.

Original worked example
Event	Included counts	Probability
Exactly 2	2	0.375
At most 2	0, 1, 2	0.875
At least 2	2, 3	0.5
From 1 through 2	1, 2	0.75

An exact count is one outcome

For three independent trials with event proportion 0.5, the probabilities for zero, one, two and three events are 0.125, 0.375, 0.375 and 0.125. Exactly two has probability 0.375. At most two includes zero, one and two, giving 0.875.

At least two includes two and three, giving 0.5. The interval from one through two includes both endpoints and has probability 0.75. Write these events separately before choosing a cumulative flag or a sequence of masses.

=BINOM.DIST(2,3,0.5,FALSE)

A finite draw changes the remaining population

Choose two records uniformly without replacement from four records, two of which are marked. Four of the six possible two-record subsets contain exactly one marked record, so the probability is 4/6. Two independent draws with replacement at proportion 0.5 instead give 0.5.

Both models have expected marked count one. Matching means do not make their count probabilities equal. A census fixes the marked count; a large sample can also force a positive minimum count.

=HYPGEOM.DIST(1,2,2,4,FALSE)

A rate belongs to a declared interval

An assumed mean of two events per hour gives mean one for a half-hour window. Under a constant-rate independent-event Poisson model, the probability of exactly one event is about 0.3678794412. Supplying mean two would answer the full-hour question.

This model assumes the event process; a single supplied mean cannot establish independence, a constant rate or a good fit to real observations.

=POISSON.DIST(1,2*0.5,FALSE)

Keep small tails visible

Subtracting a cumulative probability extremely close to one can leave numeric zero after rounding. A finite binomial or hypergeometric upper tail can instead add its individual masses.

For a positive whole threshold t, the Poisson upper tail equals GAMMA.DIST(window mean,t,1,TRUE). The incomplete-gamma identity calculates that same event directly. The course handles t=0 separately, grades nonzero results relatively and withholds models whose requested positive result is below representable floating-point precision.

=IF(B7=0,1,GAMMA.DIST(B2*B3,B7,1,TRUE))

Check the idea

For the three-trial example, which counts belong to at least two?

Try the idea

References

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