If A and B are two events such that PA - B - PA - B then show that PA - PB -2PAPBA

P(A ∪ B)=P(A) + P(B) P(A ∩ B) ⇒ #160; #160;P(A ∩ B)=P(A) + P(B) P(A ∪ B) #160;Given: P(A ∪ B) = P(A ∩ B) #160; ⇒ P(A ∩ B)=P(A) + P( ...

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Standard XII

Mathematics

Axiomatic Approach

If A and ...

Question

If $A$ and $B$ are two events such that $P (A \cup B) = P (A \cap B)$ then show that $P (A) + P (B) = 2 P (A) P (\frac{B}{A})$

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Solution

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$\Rightarrow P (A \cap B) = P (A) + P (B) - P (A \cup B)$
Given: $P (A \cup B) = P (A \cap B)$
$\Rightarrow P (A \cap B) = P (A) + P (B) - P (A \cap B)$
$\Rightarrow 2 P (A \cap B) = P (A) + P (B)$
$\Rightarrow 2 P (A \cap B) = 2 P (A) P (\frac{B}{A})$
$\Rightarrow 2 P (A \cap B) = P (A) + P (B)$

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If A and B are two events such that P(A∪B)=P(A∩B) then show that P(A)+P(B)=2P(A)P(BA)

P(A∪B)=P(A)+P(B)−P(A∩B)⇒P(A∩B)=P(A)+P(B)−P(A∪B) Given:P(A∪B)=P(A∩B) ⇒P(A∩B)=P(A)+P(B)−P(A∩B)⇒2P(A∩B)=P(A)+P(B)⇒2P(A∩B)=2P(A)P(BA)⇒2P(A∩B)=P(A)+P(B)

If $A$ and $B$ are two events such that $P (A \cup B) = P (A \cap B)$ then show that $P (A) + P (B) = 2 P (A) P (\frac{B}{A})$