Problem $42$ , chapter 4 from Introduction to Probability

An urn contains red, green, and blue balls. Balls are chosen randomly with replace- ment (each time, the color is noted and then the ball is put back). Let $r$, $g$, $b$ be the probabilities of drawing a red, green, blue ball, respectively ($r + g + b = 1$).

Find the expected number of different colors of balls obtained before getting the first red ball.

Attempt at a solution

Letting $E$ represent the expected value,


where $P(x)$ is the probability of obtaining exactly x different colours before drawing the first red ball.


$P(1)=r(b+b^2+b^3...)+r(g+g^2+g^3...)= \frac {rb}{1-b}+ \frac {rg}{1-g}$ since any number of balls of either of the other colors could be drawn in this case before the first red ball

$P(2)=1-P(1)-P(0) =(1-r-rb/(1-b)-rg/(1-g))$

So, $E= 2(1-r)-br/(r+g)-gr/(r+b)$

Is my approach correct?

  • $\begingroup$ Check the geometric distribution. $\endgroup$ Jul 30, 2020 at 3:37
  • $\begingroup$ Please surround your math with $ signs to enable MathJax $\endgroup$
    – Ryan
    Jul 30, 2020 at 3:43

1 Answer 1


This is correct, but I think the algebra obscures what is going on. For example, you might find it difficult to extend your method to $n$ colors, instead of just $3$ colors.

As an alternative method, use indicator variables.

Let $I_B$ be the indicator variable for blue (so $I_B=1$ if a blue is chosen before the first red and $0$ otherwise) and let $I_G$ be the indicator variable for green. Then $$E\left[ I_B\right]=\frac b{r+b}\quad \&\quad E\left[ I_G\right]=\frac g{r+g}$$ so the desired result is the sum $$\boxed {\frac b{r+b}+\frac g{r+g}}$$

I'll leave it as an exercise to verify that this is equivalent to your answer.


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