If P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(AB) | vedclass

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(D) The inclusion-exclusion principle states that $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)$.Substituting the given va

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None of these

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(D) The inclusion-exclusion principle states that $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)$.Substituting the given values: $P(A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.08 - x - 0.28 + 0.09 = 1.23 - x$.Given $P(A \cup B \cup C) \ge 0.75$,we have $1.23 - x \ge 0.75$,which implies $x \le 0.48$.Also,for any events $B$ and $C$,$P(BC) \ge P(ABC) = 0.09$.Furthermore,$P(B \cup C) = P(B) + P(C) - P(BC) = 0.4 + 0.8 - x = 1.2 - x \le 1$,so $x \ge 0.2$.Additionally,$P(BC) \ge P(ABC) + P(AB) - P(A) = 0.09 + 0.08 - 0.3 = -0.13$ (always true).However,the condition $P(BC) \ge P(ABC) + P(AC) - P(A) = 0.09 + 0.28 - 0.3 = 0.07$ must hold.Also,$P(B \cap C \cap A^c) = P(BC) - P(ABC) = x - 0.09 \ge 0$,so $x \ge 0.09$.Since $P(B \cup C) \le 1$,$x \ge 0.2$. Combining these,$0.2 \le x \le 0.48$. None of the options match this range.

If $P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(AB) = 0.08, P(AC) = 0.28, P(ABC) = 0.09, P(A \cup B \cup C) \ge 0.75$ and $P(BC) = x$,then find the range of $x$.

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