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.

  • 2
    $\begingroup$ It is much easier to use the converse probability: probability not seeing all three colors is equal to One minus the probability seeing all three colors. $\endgroup$ – callculus Jan 26 at 21:31
  • $\begingroup$ I have to use inclusion-exclusion; this is a multi-part problem and it specifies doing it this way $\endgroup$ – jerrythesphere Jan 26 at 22:04
  • $\begingroup$ You have to show some effort. What result do you get if you follow my advice? $\endgroup$ – callculus Jan 26 at 22:07
  • $\begingroup$ OK. Now it´s clear that you have to use the inclusion-exclusion principle. But with my advice you can double check the result which you get with the inclusion-exclusion principle. You just have to start to do something. $\endgroup$ – callculus Jan 26 at 22:19

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.

  • $\begingroup$ I understand that P(W), P(G) and P(R) represent the probabilities that white, red and green do not show up, but what would P(W $\cup$ R) represent? I assume it would be the outcomes that W and R have in common meaning only having green appear, so P(W $\cup$ R) = $\frac{2*2}{4*4}$ $\endgroup$ – jerrythesphere Jan 27 at 17:33
  • $\begingroup$ @jerrythesphere For $P(W\cap R)$ you just have to multiply the probabilities that the i-th draw is green. The probablitly to pick a green marble at the i-th draw is $P(g)=\frac{1}{1+1+2}=\frac14$. Similar for $P(w)$ and $P(r)$. The sum of these probabilities has to be $1$. $\endgroup$ – callculus Jan 27 at 18:12

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)$$


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:

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


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