I have a population that a sample was taken from, and one "group" was not selected in the sample. I am wanting to find the probability of one group not being selected. I will use red/blue balls as the objects for simplicity.

Population: 802 balls total, 37 of which are blue and the rest (765) are non-blue. A sample of 119 balls was taken without replacement (the sample can only have unique values). What is the probability that no blue balls were taken?

I think the solution is:

1 - [ 37/802 * [ (765!-(765-118)!) / (801!-(801-118)!) ] ]

1 - [ P(selecting blue) * P(selecting not blue 118 times) ]

But my solution is too big to calculate. Is this correct, and what is an easier solution?


1 Answer 1


This "without replacement" sample has a hypergeometric distribution, but it may be more obvious if you say there are ${765 \choose 119}$ ways of choosing all non-blues (ignoring order) out of ${802 \choose 119}$ total equally likely ways.

Or, if you want to take order into account, then as $\frac{765}{802}\times\frac{764}{801}\times\cdots \times \frac{647}{683}$ or equivalently $\frac{765!}{(765 -119)!}$ out of $\frac{802!}{(802 -119)!}$ ways. You will get the same result.

These are still big numbers, and you may want a computer to help you; logarithms might help in some cases. With R there are several possibilities, including

dhyper(0, 37, 765, 119)


exp(lchoose(765,119) - lchoose(802,119))

both of which give about $0.00226$. So seeing no blues is unlikely.

The probability would be different if you had several groups and would be surprised if any group was missed, though the calculation would be more complicated. Suppose you had $22$ colours with $36$ or $37$ of each giving $802$ in total. Then simulation suggests to me the probability of not seeing all $22$ colours might be about $0.054$, which is much less unlikely.


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