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Question

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')

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Answered 2 years ago
Answered 2 years ago
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Given events A,B,CA, B, C such that:

AB=, AC=, BC=P(A)=0.2P(B)=0.3P(C)=0.4\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,CA;B,C are mutually exclusive.

a) P(ABC)P(A\cap B\cap C)

Use formula (2-4) - $\textit{For mutually exclusive events E1,E2,EnE_1, E_2, \dotsc E_n:}$

P(E1E2En)=P(E1)+P(E2)++P(En)\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;CA;B;C are mutually exclusive, direct use of this formula and substitution of values given above:

P(ABC)=P(A)+P(B)+P(C)=0.2+0.3+0.4=0.9\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(ABC)P(A\cap B\cap C)

ABC=A\cap B\cap C=\varnothing

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

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

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

c) P(AB)P(A\cap B)

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

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

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