Let A and B be two independent events. The pr | vedclass

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(B) Given $P(A \cap B) = \frac{1}{6}$ and $P(A^c \cap B^c) = \frac{1}{3}$.Since $P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B) = \frac{1}{3}$,we

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$\frac{1}{2}$ or $\frac{1}{3}$

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(B) Given $P(A \cap B) = \frac{1}{6}$ and $P(A^c \cap B^c) = \frac{1}{3}$.Since $P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B) = \frac{1}{3}$,we have $P(A \cup B) = \frac{2}{3}$.Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{3}$.Thus,$P(A) + P(B) = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}$.Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B) = \frac{1}{6}$.Let $x = P(A)$ and $y = P(B)$. Then $x + y = \frac{5}{6}$ and $xy = \frac{1}{6}$.These are roots of the quadratic equation $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{5}{6}t + \frac{1}{6} = 0$.Multiplying by $6$,we get $6t^2 - 5t + 1 = 0$.$(2t - 1)(3t - 1) = 0$,so $t = \frac{1}{2}$ or $t = \frac{1}{3}$.Therefore,$P(A)$ is $\frac{1}{2}$ or $\frac{1}{3}$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

Let $A$ and $B$ be two independent events. The probability that both $A$ and $B$ occur together is $1/6$ and the probability that neither of them occurs is $1/3$. The probability of occurrence of $A$ is

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