# Inclusion Exclusion Principle (language confusion)

Suppose that n independent trials, each of which results in any of the outcomes $0, 1, 2$, with respective probabilities $0.3, 0.5$, and $0.2$, are performed. Find the probability that both outcome $1$ and outcome $2$ occur at least once.

I tried finding the complementary event but I'm stuck at understanding the language involved. Is it both B and C doesn't occur anytime or both B and C doesn't occur together?

Am I right if I say that the answer's just $1 - Pr(A \ \textrm{occurs}) = 1 - 0.3^n$

We have that $$P(\mbox{1 and 2 occur at least once})=1-P(\mbox{(1 never occurs) or (2 never occurs)})$$ $$P(\mbox{(1 never occurs) or (2 never occurs)})=P(\mbox{1 never occurs})+P(\mbox{2 never occurs})-P(\mbox{1 and 2 never occur}).$$ Now $$P(\mbox{1 never occurs})=P(\mbox{0 or 2 always occur})=(0.3+0.2)^n=(0.5)^n,$$ $$P(\mbox{2 never occurs})=P(\mbox{0 or 1 always occur})=(0.3+0.5)^n=(0.8)^n,$$ $$P(\mbox{1 and 2 never occur})=P(\mbox{0 always occurs})=(0.3)^n.$$ Finally, we obtain $$P(\mbox{1 and 2 occur at least once})=1-(0.5)^n-(0.8)^n+(0.3)^n.$$

• Why doesn't the event 0 occurs or not matters in this case? I mean it doesn't occurs in any of the equations you wrote – uzumaki Oct 1 '16 at 18:13
• @uzumaki Is it better now? – Robert Z Oct 1 '16 at 18:17
• but a little doubt P(1 and 2 never occurs) = P(1 never occurs)*P(2 never occurs) = 0.4^n as the events are independent what's wrong in this step? – uzumaki Oct 1 '16 at 18:26
• @uzumaki (1 and 2 never occur) means always 3. (1 never occurs) and (2 never occurs) are not independent. – Robert Z Oct 1 '16 at 18:34

Using the De Morgan's low we have tha the complementary of $B$ and $C$ is not $B$ or not $C$. So therefore you need to find the cases when only $A$ and $B$ occurs, as well as only $A$ and $C$ and then subtract these probability from $1$ to get the wanted probability.

• Wouldn't it be difficult to calculate as A and B or A and C can occur many different time during the n trials? Also why are we ignoring only A occurs in this – uzumaki Oct 1 '16 at 18:04

I will first try to clarify the problem's meaning by formalizing it. Formal language is ideal for clarifying ambiguities. I will then proceed to solving it.

Step 1: Formalizing

Let $(\Omega, \mathcal{F}, P)$ be the underlying probability space, and let $X_1, \dots, X_n : \Omega \rightarrow \mathbb{R}$ be random variables denoting the trial outcomes, respectively. It is given that $X_1, \dots, X_n$ are i.i.d. (w.r.t. $P$). Moreover, it is given that \begin{align} P(X_1 = 0) &= 0.3, \\ P(X_1 = 1) &= 0.5, \\ P(X_1 = 2) &= 0.2. \end{align}

It is required to compute the probability $P(A_1 \cap A_2)$, where $A_1, A_2$ are the events \begin{align} A_1 &:= \{X_1 = 1\} \cup \cdots \cup \{X_n = 1\}, \\ A_2 &:= \{X_1 = 2\} \cup \cdots \cup \{X_n = 2\}. \end{align}

Step 2: Solving

Taking complements w.r.t. $\Omega$, we have \begin{align} P(A_1 \cap A_2) &= 1-P\left(\overline{A_1 \cap A_2}\right) \\ &\overset{\text{De Morgan}}{=} 1-P\left(\overline{A_1}\cup\overline{A_2}\right) \\ &\overset{\text{incl. excl.}}{=} 1 - \left(P\left(\overline{A_1}\right) + P\left(\overline{A_2}\right) - P\left(\overline{A_1}\cap\overline{A_2}\right)\right). \tag{1}\label{eq1} \end{align}

Now, \begin{align} P\left(\overline{A_1}\right) &\overset{\text{De Morgan}}{=} P\left(\overline{\{X_1=1\}}\cap\cdots\cap\overline{\{X_n = 1\}}\right) \\ &\overset{\text{indep.}}{=} P\left(\overline{\{X_1 = 1\}}\right) \cdots P\left(\overline{\{X_n = 1\}}\right) \\ &\overset{\text{ident. dist.}}{=} \left(P\left(\overline{\{X_1 = 1\}}\right)\right)^n \\ &= \left(1-P(X_1 = 1)\right)^n \\ &= (1-0.5)^n \\ &= 0.5^n. \tag{2}\label{eq2} \end{align}

Similarly, $$P\left(\overline{A_2}\right) = \left(1-P(X_1 = 2)\right)^n = \left(1-0.2\right)^n = 0.8^n. \tag{3}\label{eq3}$$

Finally, observing that \begin{align} \overline{A_1}\cap\overline{A_2} &= \left(\{X_1 \neq 1\}\cap\cdots\cap\{X_n \neq 1\}\right) \cap \left(\{X_1 \neq 2\}\cap\cdots\cap\{X_n \neq 2\}\right) \\ &= \left(\{X_1 \neq 1\}\cap \{X_1 \neq 2\}\right)\cap \cdots \cap\left(\{X_n \neq 1\}\cap \{X_n \neq 2\}\right) \\ &= \{X_1 = 0\} \cap \cdots \cap \{X_n = 0\}, \end{align} we have \begin{align} P\left(\overline{A_1}\cap\overline{A_2}\right) &= P\left(\{X_1 = 0\}\cap\cdots\cap\{X_n = 0\}\right) \\ &\overset{\text{indep.}}{=} P(X_1 = 0) \cdots P(X_n = 0) \\ &\overset{\text{ident. dist.}}{=} \left(P(X_1 = 0)\right)^n \\ &= 0.3^n. \tag{4}\label{eq4} \end{align}

Substituting \eqref{eq2}, \eqref{eq3} and \eqref{eq4} into \eqref{eq1}, we obtain $$P(A_1 \cap A_2) = 1-0.5^n-0.8^n+0.3^n.$$

$A=$ the case that there are no $1$s. $P(A)=0.5^n$
$B=$ the case that there are no $2$s. $P(B)=0.8^n$
$A\cap B=$ the case there are no $1$s or $2$s. $P(A\cap B)=0.3^n$

Inclusion-Exclusion says that the probability there are no $1$s or no $2$s is $$P(A)+P(B)-P(A\cap B)=0.5^n+0.8^n-0.3^n\tag{1}$$ That means that the probability that there is at least one of each is $$\bbox[5px,border:2px solid #C0A000]{1-0.5^n-0.8^n+0.3^n}\tag{2}$$ Note that to get both a $1$ and a $2$, we will need at least $2$ trials. If $n=0$ or $n=1$, $(2)$ gives a probability of $0$. If $n=2$, we get a probability of $0.2$, which is the probability of getting a $1$ then a $2$ or getting a $2$ then a $1$.