Problem 7.19 Easy Free 1 variation

Last Green Ball on a Given Draw

A bin contains 4 green balls and 11 red balls. Balls are drawn one at a time without replacement until all are removed. Find the probability that the 4th (last) green ball is drawn exactly on the 9th draw.

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. All $\binom{15}{4}$ choices of the four positions (out of the 15 draws) occupied by green balls are equally likely. This holds because the drawing order is a uniformly random permutation of the balls.

The last green ball is drawn on draw 9 exactly when two things happen:

draw 9 is green;
exactly three of draws $1,\dots,8$ are green.

The remaining positions $10,\dots,15$ are then automatically red. Such position sets are determined by choosing the three earlier green positions among the first 8, which can be done in $\binom{8}{3}=56$ ways. Hence

\mathbb{P}(\text{last green on draw }9)=\frac{\binom{8}{3}}{\binom{15}{4}}=\frac{56}{1365}=\frac{8}{195}\approx 0.041.

\boxed{\dfrac{8}{195}}

Generalisation. With $g$ green balls among $n$ in total, the last green ball is drawn on draw $k$ with probability $\binom{k-1}{g-1}\big/\binom{n}{g}$ .

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 7.19a

A box contains 3 shiny pennies and 4 dull pennies. Pennies are drawn one at a time without replacement. Find the probability that more than 4 draws are needed to see all 3 shiny pennies.

Work it out, then compare with the answer.

Solution

The 3 shiny positions form a uniformly random 3-subset of the 7 draws. At most 4 draws are needed exactly when all three shiny positions lie in $\{1,2,3,4\}$ . This happens with probability

\frac{\binom{4}{3}}{\binom{7}{3}}=\frac{4}{35}.

The complement gives

\mathbb{P}(\text{more than 4 draws needed})=1-\frac{4}{35}=\frac{31}{35}.

\boxed{\dfrac{31}{35}}

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