If P(A) = frac{3}{10},P(B) = frac{3}{5},and P | vedclass

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(B) We are given $P(A) = \frac{3}{10}$,$P(B) = \frac{3}{5}$,and $P(A \cup B) = \frac{3}{5}$.Using the addition theorem of probability: $P(A \cup B) =

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$\frac{1}{12}$

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(B) We are given $P(A) = \frac{3}{10}$,$P(B) = \frac{3}{5}$,and $P(A \cup B) = \frac{3}{5}$.Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values: $\frac{3}{5} = \frac{3}{10} + \frac{3}{5} - P(A \cap B)$.This simplifies to $P(A \cap B) = \frac{3}{10}$.Now,we need to calculate $P(A|B) 	imes P(B|A)$.By definition,$P(A|B) = \frac{P(A \cap B)}{P(B)}$ and $P(B|A) = \frac{P(A \cap B)}{P(A)}$.Thus,$P(A|B) 	imes P(B|A) = \frac{P(A \cap B)}{P(B)} 	imes \frac{P(A \cap B)}{P(A)} = \frac{(\frac{3}{10})}{(\frac{3}{5})} 	imes \frac{(\frac{3}{10})}{(\frac{3}{10})} = \frac{1}{2} 	imes 1 = \frac{1}{2}$.Wait,re-evaluating the calculation: $P(A \cap B) = \frac{3}{10} + \frac{3}{5} - \frac{3}{5} = \frac{3}{10}$.$P(A|B) = \frac{3/10}{3/5} = \frac{1}{2}$.$P(B|A) = \frac{3/10}{3/10} = 1$.Result is $\frac{1}{2} 	imes 1 = \frac{1}{2}$.Given the options,let's re-check the input values. If $P(B) = 2/5$ was intended,the result would be $1/12$. Based on the provided values,the calculation is $1/2$.

If $P(A) = \frac{3}{10}$,$P(B) = \frac{3}{5}$,and $P(A \cup B) = \frac{3}{5}$,then find the value of $P(A|B) \times P(B|A)$.

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