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If P and Q are two random events, then the following is TRUE

Option: 1

Independence of P and Q implies that probability \mathrm{(P \cap Q)=0}


Option: 2

\mathrm{Probability (P \cup Q) \geq Probability (P)+ Probability (Q)}


Option: 3

If P and Q are mutually exclusive, then they must be independent


Option: 4

\mathrm{\text { Probability }(P \cap Q) \leq \text { Probability }(P)}


Answers (1)

best_answer

If A and B are disjoint (or mutually exclusive means \mathrm{A \cap B=\phi}) events then.

            \mathrm{P(A \cup B)=P(A)+P(B)}

Or        \mathrm{P(A \cap B)=0}

For any events A and B,

          \begin{aligned} \mathrm{P(A \cup B)}= & \mathrm{P(A)+P(B)-P(A \cap B) \leq P(A)} \\ & \mathrm{+P(B)} \end{aligned}

If A and B are independent events then

          \begin{aligned} \mathrm{P(A \cap B)}= & \mathrm{P(A) \times P(B)} \\ & \mathrm{P(A \cap B) \leq\{P(A), P(B)\}} \end{aligned}

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HARSH KANKARIA

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