Question
If $A$ and $B$ are any two events, the probability that exactly one of them occurs is(A) $P(A)+P(B)-2 P(A \cap B)$(B) $P(\bar{A})+P(\bar{B})-2 P(\bar{A} \cap \bar{B})$(C) $P(A \cap(\bar{B}))+P((\bar{A}) \cap B)$(D) $P(A)+P(B)-P(A \cup B)$
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This can be mathematically represented as: \[P(A \cap \bar{B}) + P(\bar{A} \cap B)\] Show more…
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$A$ and $B$ are two events, The probability that at most one of $A, B$ occurs is (a) $1-P(A \cap B)$ (b) $P\left(A^{\prime}\right)+P\left(B^{\prime}\right)-P\left(A^{\prime} \cap B^{\prime}\right)$ (c) $P\left(A^{\prime}\right)+P\left(B^{\prime}\right)+P(A \cup B)-1$ (d) $P\left(A \cap B^{\prime}\right)+P\left(A^{\prime} \cap B\right)+P\left(A^{*} \cap B^{\prime}\right)$.
If two events, A and B are mutually exclusive, then Choose one or more. a. Their intersection is equal to the product of their probabilities (ie. P(A and B)=P(A)P(B)) b. Their union is the sum of individual probabilities, P(A)+P(B). c. The events don't intersect. d. Their intersection is equal to the conditional probability of B given A.
The event that ${ }^{*} A$ or $B$ but not both" will occur can be written as $$ \left(A \cap B^{\prime}\right) \cup\left(A^{\prime} \cap B\right) $$ Express the probability of this event in terms of $P(A)$ $P(B)$, and $P(A \cap B)$
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