Last green ball appearing on the ninth draw

Question

A bin contains $4$ green and $11$ red balls, drawn out one by one at random without replacement. What is the probability that the ninth draw is the one that removes the final green ball?

Short answer

$8/195$ . All $\binom{15}{4}$ placements of the green balls in the draw order are equally likely. We need one green in position $9$ and the other three among positions $1$ – $8$ : $\binom{8}{3}/\binom{15}{4}=56/1365=8/195$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Counting and Combinatorics
Sample Spaces, Events, and the Probability Axioms
Conditional Probability, Total Probability, and Bayes' Rule

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