If $A$ and $B$ are mutually exclusive events with $P(A) = \frac{1}{2} imes P(B)and\ A \cup B = S$,(total sample space) then $P(A)$ is equal to

Question

If $A$ and $B$ are mutually exclusive events with $P(A) = \frac{1}{2} 	imes P(B)and\ A \cup B = S$,(total sample space) then $P(A)$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given, $A \cup B = S$
$	herefore P(A \cup B) = P(S) = 1$
$\Rightarrow P(A) + P(B) = 1\ \ \ \ \ \lbrack\because P(A \cap B) = 0brack$
$\Rightarrow P(A) + 2P(A) = 1$
$\Rightarrow P(A) = \frac{1}{3}\ \ \ \ \lbrack\because P(B) = 2P(A),\ givenbrack$

Q. If $A$ and $B$ are mutually exclusive events with $P(A) = \frac{1}{2} \times P(B)and\ A \cup B = S$,(total sample space) then $P(A)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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