Question
Let $A$ and $B$ be two events such that $P\left((A \cup B)^{C}\right)=0.6$ and $P(A \cap B)=0.1$. Let $E$ be the event that either $A$ or $B$ but not both will occur. Find $P(E \mid \Lambda \cup B)$.
Step 1
6$. This means that the probability of neither $A$ nor $B$ occurring is $0.6$. Therefore, the probability of either $A$ or $B$ occurring (or both) is $1 - 0.6 = 0.4$. This is the probability of $A \cup B$. Show more…
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