If A and B are two events such that P(A) = 0. | vedclass

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(A) Given: $P(A) = 0.7$,$P(B) = 0.4$,$P(A \cap \overline{B}) = 0.5$. First,find $P(A \cap B)$: $P(A) = P(A \cap B) + P(A \cap \overline{B}) \implies 0

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) Given: $P(A) = 0.7$,$P(B) = 0.4$,$P(A \cap \overline{B}) = 0.5$. First,find $P(A \cap B)$: $P(A) = P(A \cap B) + P(A \cap \overline{B}) \implies 0.7 = P(A \cap B) + 0.5 \implies P(A \cap B) = 0.2$. Next,find $P(A \cup \overline{B})$: $P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B}) = 0.7 + (1 - 0.4) - 0.5 = 0.7 + 0.6 - 0.5 = 0.8$. Now,calculate $P(B \mid (A \cup \overline{B}))$: $P(B \mid (A \cup \overline{B})) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})} = \frac{P((B \cap A) \cup (B \cap \overline{B}))}{P(A \cup \overline{B})} = \frac{P(A \cap B) + 0}{0.8} = \frac{0.2}{0.8} = \frac{1}{4}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

If $A$ and $B$ are two events such that $P(A) = 0.7$,$P(B) = 0.4$,and $P(A \cap \overline{B}) = 0.5$,where $\overline{B}$ denotes the complement of $B$,then $P(B \mid (A \cup \overline{B}))$ is equal to:

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