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$\def\Prob{\mathop{\rm Prob}}$

For events $A, B$ from a sample space $X$, label $A^\complement$ the event complementary to $A$ (i.e., $A \cup A^\complement = X, A \cap A^\complement = \emptyset$ ), and $\Prob[A \mid B]$ denotes the probability of $A$, given that B has occurred.

1) Go with two events $A, B$ from a sample space $X$, for which $\Prob[A] = 1/6$ , and $\Prob[A ∪ B] = 2/3$ . If $A$ and $B$ are independent events, calculate the following probabilities:

a) $\Prob[B] =$

b) $\Prob[A \cap B] =$

c) $\Prob[A^\complement] =$

d) $\Prob[B \mid A] =$

I am stuck on how to even do this- any hint to get started will be greatly appreciated. trying to understand it not only looking for answers.

thanks.

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2/3 = Prob[A ∪ B] = Prob(A) + Prob(B and !A) = Prob(A) + Prob(B) * (1 - Prob(A)) = 1/6 + Prob(B) * (1 - 1/6)

Now finding Prob(B) is straightforward algebra.

The other probabilities just require restating them as expressions involving the probabilities of A and B.

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  • $\begingroup$ i got Prob(B) = 1/10... did i do the math right? $\endgroup$ – carlton.x Mar 11 '17 at 18:21
  • $\begingroup$ @carlton.x I get 2/3 = 1/6 + Prob(B)*(1-1/6) so 1/2 = Prob(B)(5/6) so 6/10 = Prob(B). $\endgroup$ – btilly Mar 12 '17 at 1:48
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Independence means: $\Prob(A\cap B)=\Prob (A)\cdot\Prob(B)$

Also recall: $\Prob(A\cup B)=\Prob(A)+\Prob(B)-\Prob(A\cap B)$

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