If 6 P(A) = 8 P(B) = 14 P(A cap B) = 1,then P | vedclass

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(A) Given that $6 P(A) = 1 \implies P(A) = \frac{1}{6}$.Given that $8 P(B) = 1 \implies P(B) = \frac{1}{8}$.Given that $14 P(A \cap B) = 1 \implies P(

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$\frac{3}{7}$

Vedclass · Answer

(A) Given that $6 P(A) = 1 \implies P(A) = \frac{1}{6}$.Given that $8 P(B) = 1 \implies P(B) = \frac{1}{8}$.Given that $14 P(A \cap B) = 1 \implies P(A \cap B) = \frac{1}{14}$.We need to find $P(A' \mid B)$.Using the conditional probability formula,$P(A' \mid B) = \frac{P(A' \cap B)}{P(B)}$.We know that $P(A' \cap B) = P(B) - P(A \cap B)$.Substituting the values: $P(A' \cap B) = \frac{1}{8} - \frac{1}{14} = \frac{7 - 4}{56} = \frac{3}{56}$.Now,$P(A' \mid B) = \frac{3/56}{1/8} = \frac{3}{56} 	imes 8 = \frac{3}{7}$.Thus,the correct option is $A$.

If $6 P(A) = 8 P(B) = 14 P(A \cap B) = 1$,then $P(A' \mid B) = $ . . . . . . .

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