# Probability - Urn with two red and one black balls

Consider an urn with two red and one black balls. Draw four times with replacement.

a) What is the probability that exactly one black ball will be drawn?

b) What is the probability that at least one black ball will be drawn?

a) Since there are $$4$$ possible ways to draw exactly one black ball:

b r r r
r b r r
r r b r
r r r b


Where b stands for black and r for red. If we're drawing with replacement, we have $$3^4 = 81$$ possible outcomes and hence the probability that exactly one black ball will be drawn $$= \dfrac{4}{81}.$$

b) With the same approach taken in (a), we would have:

$$1$$ way of drawing $$4$$ black balls.

$$4$$ ways of drawing $$3$$ black balls and $$1$$ red one.

$$4$$ ways of drawing $$2$$ black balls and $$2$$ red ones.

$$4$$ ways of drawing $$1$$ black ball and $$3$$ red ones.

Thus the probability that at least one black ball will be drawn $$= \dfrac{13}{81}.$$

However, my textbook gives (a) $$\dfrac{32}{81}$$ and (b) $$\dfrac{65}{81}$$ as solutions. What's wrong with my reasoning?

• The probability of drawing a black ball on each selection is $1/3$, while the probability of drawing a red ball on each selection is $2/3$. You have to take these probabilities into account. Commented Sep 18, 2020 at 22:07
• You have to view it as 4 independent experiments in each of which we draw only one ball. Then we can multiply the probabilities of the individual events. Commented Sep 18, 2020 at 22:09
• Since the selections are made with replacement, this is a binomial distribution problem. Commented Sep 18, 2020 at 22:09
• If you are looking for a combinatorial solution, you should take the balls as distinct, not identical. Commented Sep 18, 2020 at 23:29

Exactly one black ball: we have 4 ways to pick the unique experiment that will have a black ball, as you already noted. In any experiment we have a chance of $$\frac{1}{3}$$ of a black ball and $$\frac{2}{3}$$ of a red ball. So any sequence of outcomes like $$rrrb$$ or $$brrr$$ has total probability of $$\frac{1}{3} \times (\frac{2}{3})^3 = \frac{8}{81}$$ and as we have four mutually exclusive such sequences we have $$\frac{32}{81}$$ for the first as your textbook says.
For "at least one black ball" it's easier to count the complementary event: only red balls. This happens in a unique sequence with chance $$(\frac{2}{3})^4 = \frac{16}{81}$$. So our event happens with chance $$1 - \frac{16}{81} = \frac{65}{81}$$
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