For two events A and B,P(A)=0.5,P(A cup B)=0. | vedclass

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(B) Given: $P(A)=0.5$,$P(A \cup B)=0.6$,and $P(B)=K$. Since $A$ and $B$ are independent events,the probability of their intersection is given by $P(A

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$0.2$

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(B) Given: $P(A)=0.5$,$P(A \cup B)=0.6$,and $P(B)=K$. Since $A$ and $B$ are independent events,the probability of their intersection is given by $P(A \cap B) = P(A) 	imes P(B) = 0.5K$. Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Substituting the given values: $0.6 = 0.5 + K - 0.5K$. Simplifying the equation: $0.6 = 0.5 + 0.5K$. Subtracting $0.5$ from both sides: $0.1 = 0.5K$. Solving for $K$: $K = \frac{0.1}{0.5} = 0.2$.

$P(A)$	$P(B)$	$P(A \cap B)$	$P(A \cup B)$
$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	$........$

For two events $A$ and $B$,$P(A)=0.5$,$P(A \cup B)=0.6$ and $P(B)=K$ are given. If $A$ and $B$ are independent events,then $K=$ . . . . . . .

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