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If $$A, B,$$ and $$C$$ are mutually exclusive events with
$$P(A)=0.2, P(B)=0.3,$$ and $$P(C)=0.4,$$ determine the fol-
lowing probabilities:

$$\begin{array}{ll}{\text { (a) } P(A \cup B \cup C)} & {\text { (b) } P(A \cap B \cap C)} \\ {\text { (c) } P(A \cap B)} & {\text { (d) } P[(A \cup B) \cap C]} \\ {\text { (e) } P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)}\end{array} $$

$$A, B, C$$

$$P(A)=0.2 \quad P(B)=0.3 \quad P(c)=0.4$$

a) $$P(A \cup B \cup C)=P(A)+P(B)+P(C)$$

$$=0.2+0.3+0.4=0.9$$

$$P(A \cap B \cap C)=0$$

c) $$P(A \cap B) \rightarrow A \cap B=\phi$$

$$\Rightarrow P(A \cap B)=0$$

d) $$P[(A \cup B) \cap C]=0$$

e) $$P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)$$

$$=1-[P(A)+P(B)+P(C)]$$

$$=1-[0.2+0.3+0.4]=0.1$$

 

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