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EEJ MAIN Mathematics QUESTION #6979
Question 1

For three events $A$, $B$, $C$:
$P(\text{exactly one of }A\text{ or }B) = P(\text{exactly one of }B\text{ or }C) = P(\text{exactly one of }C\text{ or }A) = \dfrac{1}{4}$
$P(A\cap B\cap C) = \dfrac{1}{16}$.
Find $P(\text{at least one of }A, B, C\text{ occurs})$.

  • $\dfrac{7}{32}$
  • $\dfrac{7}{16}$✔️
  • $\dfrac{7}{64}$
  • $\dfrac{3}{16}$
Correct Answer Explanation

Use: $P(\text{exactly one of }X\text{ or }Y) = P(X)+P(Y)-2P(X\cap Y)$.

Adding all three equations:

$2[P(A)+P(B)+P(C)] - 2[P(A\cap B)+P(B\cap C)+P(C\cap A)] = \dfrac{3}{4}$

$\Rightarrow P(A)+P(B)+P(C) - [P(A\cap B)+P(B\cap C)+P(C\cap A)] = \dfrac{3}{8}$ ...(i)

Using inclusion-exclusion:

$P(A\cup B\cup C) = P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$

$= \dfrac{3}{8} + \dfrac{1}{16} = \dfrac{6}{16}+\dfrac{1}{16} = \dfrac{7}{16}$