Probability Inclusion-Exclusion With 3 Events

Question

The question is: An urn contains 4 balls: 1 white, 1 green, and 2 red. We draw 3 balls with replacement. Find the probability we did not see all three colors.

I need to define the events as W= {white ball did not appear} and similarly for R and G, while specifically using inclusion-exclusion to solve the problem.

My first thought was to use the identity P(A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B) and solve for (A $\cap$ B) but I would only be able to use it for two colors at a time.

Answer 1

Here is how the principle of inclusion-exclusion looks with three events:

$\begin{align*} \Pr(W\cup R\cup G) &= \Pr(W)+ \Pr(R)+ \Pr(G)\\ &\quad-\Pr(W\cap R)-\Pr(W\cap G)-\Pr(G\cap R)\\ &\qquad+ \Pr(W\cap R\cap G) \end{align*}$

It’s up to you to compute each of the terms on the RHS.

Answer 2

The principle of inclusion and exclusion works for any number of events:

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$$

Answer 3

Let $W,R,G$ the events, that the white, red and green marbles do not show up.

Then the number of combinations for W are:

$a) \ ggg\rightarrow 1, b) \ ggr\rightarrow 3, c) \ rrg\rightarrow 3, d)\ rrr\rightarrow 1$

The corresponding probabilities are

$a) \ \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4}= \frac{1}{64}$

$b) \ 3\cdot \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{2}= \frac{3}{32}$

$c) \ 3\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{4}= \frac{3}{16}$

$d) \ \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}= \frac{1}{8}$

Therefore $P(W)=\frac{27}{64}$

Now what ist $P(W\cap R)$?

$W\cap R$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $ ggg$

$P(W\cap R)=\frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4}= \frac{1}{64}$

You can check your final result by using the converse probability:

$P(\textrm{"Do not see all three colors"})=1-P(\textrm{"See all three colors"})$