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?
Here's my answers:
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?
Best Answer
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 $$1 - \frac{16}{81} = \frac{65}{81} $$
again as your book says.