## Related questions with answers

If A, B, and C are mutually exclusive events with P(A) = 0.2, P(B) = 0.3, P(C) = 0.4, determine the following probabilities. a. P(A ∪ B ∪ C) b. P(A ∩ B ∩ C) c. P(A ∩ B) d. P[A ∪ B) ∩ C] e. P(A' ∩ B' ∩ C')

Solution

VerifiedGiven events $A, B, C$ such that:

$\begin{gather*} A\cap B=\varnothing ,\ A\cap C=\varnothing,\ B\cap C=\varnothing\\ P(A)=0.2\\ P(B)=0.3\\ P(C)=0.4 \end{gather*}$

The first statements mean that $A;B,C$ are mutually exclusive.

#### a) $P(A\cap B\cap C)$

Use formula (2-4) - $\textit{For mutually exclusive events $E_1, E_2, \dotsc E_n$:}$

$\begin{equation*} P(E_1\cup E_2\cup \dotsc \cup E_n)=P(E_1)+P(E_2)+\dotsc + P(E_n) \end{equation*}$

$A;B;C$ are mutually exclusive, direct use of this formula and substitution of values given above:

$\begin{equation*} P(A\cup B\cup C)=P(A)+P(B)+P(C)=0.2+0.3+0.4=0.9 \end{equation*}$

#### b) $P(A\cap B\cap C)$

$A\cap B\cap C=\varnothing$

Because any element of $A\cap B\cap C$ should be in all three sets $A,B$ and $C$, but from mutual exclusiveness, we know that there are no outcomes that are in both $A$ and $B$, for example, so no outcome is in all three.

$\begin{equation*} P(A\cap B\cap C)=P(\varnothing )\overset{(A)}{=}0 \end{equation*}$

$(A):\ P(\varnothing)=0$ is proven in exercise 2.48.

#### c) $P(A\cap B)$

$A\cap B=\varnothing$ - this is a part of mutual exclusiveness:

$\begin{equation*} P(A\cap B)=P(\varnothing )\overset{(A)}{=}0 \end{equation*}$

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