(B) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Since $P(A) = 1 - P(A^C) = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting the values into the

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(B) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Since $P(A) = 1 - P(A^C) = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting the values into the addition theorem of probability:$\frac{5}{6} = \frac{1}{2} + P(B) - \frac{1}{3}$.$\frac{5}{6} = \frac{3-2}{6} + P(B) = \frac{1}{6} + P(B)$.$P(B) = \frac{5}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$.Therefore,$P(B^C) = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3}$.

Class 11 Mathematics Probability Odds in favor and odds against, Addition theorem on probability

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If $A$ and $B$ are two events with $P(A \cap B) = \frac{1}{3}$,$P(A \cup B) = \frac{5}{6}$,and $P(A^C) = \frac{1}{2}$,then the value of $P(B^C)$ is

AP EAMCET 2021 All AP EAMCET

A
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{2}{3}$
D
$\frac{5}{6}$

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Odds in favor and odds against, Addition theorem on probability

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EasyIIT 1990

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If $A$ and $B$ are two events with $P(A \cap B) = \frac{1}{3}$,$P(A \cup B) = \frac{5}{6}$,and $P(A^C) = \frac{1}{2}$,then the value of $P(B^C)$ is

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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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