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For three events $A,B $ and $C$ ,$P ($ Exactly one of $A$ or $B$ occurs$)\, =\, P ($ Exactly one of $C$ or $A$ occurs $) =$ $\frac{1}{4}$ and $P ($ All the three events occur simultaneously $) =$ $\frac{1}{16}$ Then the probability that at least one of the events occurs is :
$\frac{3}{{16}}$
$\frac{7}{{32}}$
$\frac{7}{{16}}$
$\frac{7}{{64}}$
Solution
$\mathrm{P}$ (exactly one of $A$ or $B$ occurs)
$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$ ….$(1)$
$\mathrm{P}(\text { Exactly one of } \mathrm{B} \text { or $C$ occurs })$
$=P(B)+P(C)-2 P(B \cap C)=\frac{1}{4}$ …..$(2)$
P (Exactly one of $C$ or $A$ occurs)
$=\mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A})-2 \mathrm{P}(\mathrm{C} \cap \mathrm{A})=\frac{1}{4}$ …..$(3)$
Adding $(1),(2)$ and $(3),$ we get
$2 \Sigma P(A)-2 \Sigma P(A \cap B)=\frac{3}{4}$
$\therefore \Sigma P(A)-\Sigma P(A \cap B)=\frac{3}{8}$
Now, $\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=\frac{1}{16}$
$\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})$
$=\Sigma P(A)-\Sigma P(A \cap B)+P(A \cap B \cap C)$
$=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}$
