The probabilities of three events A, B and C | vedclass

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(C) Given $P(A)=0.6, P(B)=0.4, P(C)=0.5, P(A \cup B)=0.8, P(A \cap C)=0.3, P(A \cap B \cap C)=0.2$.Using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we

Vedclass · Accepted Answer

$[0.25, 0.35]$

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(C) Given $P(A)=0.6, P(B)=0.4, P(C)=0.5, P(A \cup B)=0.8, P(A \cap C)=0.3, P(A \cap B \cap C)=0.2$.Using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get $0.8 = 0.6 + 0.4 - P(A \cap B)$,so $P(A \cap B) = 0.2$.Using the formula $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$:$\alpha = 0.6 + 0.4 + 0.5 - 0.2 - \beta - 0.3 + 0.2 = 1.2 - \beta$.Given $0.85 \leq \alpha \leq 0.95$,we have $0.85 \leq 1.2 - \beta \leq 0.95$.Subtracting $1.2$ from all parts: $-0.35 \leq -\beta \leq -0.25$.Multiplying by $-1$ reverses the inequality: $0.25 \leq \beta \leq 0.35$.Thus,$\beta \in [0.25, 0.35]$.

The probabilities of three events $A, B$ and $C$ are given by $P(A)=0.6, P(B)=0.4$ and $P(C)=0.5$. If $P(A \cup B)=0.8, P(A \cap C)=0.3, P(A \cap B \cap C)=0.2, P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$,where $0.85 \leq \alpha \leq 0.95$,then $\beta$ lies in the interval:

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