If P(A)=0.25, P(B)=0.55 and P(A cup B)=0.65,t | vedclass

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(C) Given that $P(A)=0.25, P(B)=0.55$ and $P(A \cup B)=0.65$. First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) Given that $P(A)=0.25, P(B)=0.55$ and $P(A \cup B)=0.65$. First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $0.65 = 0.25 + 0.55 - P(A \cap B)$ $0.65 = 0.80 - P(A \cap B)$ $P(A \cap B) = 0.80 - 0.65 = 0.15$. Now,we need to find $P(B' \mid A)$. Using the conditional probability formula,$P(B' \mid A) = \frac{P(B' \cap A)}{P(A)}$. Since $P(B' \cap A) = P(A) - P(A \cap B)$,we have: $P(B' \cap A) = 0.25 - 0.15 = 0.10$. Therefore,$P(B' \mid A) = \frac{0.10}{0.25} = \frac{10}{25} = 0.4$. Thus,the correct option is $C$.

If $P(A)=0.25, P(B)=0.55$ and $P(A \cup B)=0.65$,then $P(B' \mid A) =$ . . . . . . .

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