Question 1190899
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I need your help with this!
Assume Pr[A]=0.3, Pr[B]=0.4, and Pr[A∩B]=0.05. Find the following values:
(a)   Pr[B|A] =
(b)   Pr[A|B′] =
(c)   Pr[B′|A′] =
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            In this problem,  all questions ask about  CONDITIONAL  PROBABILITY.


            To solve the problem and to answer all the questions,  you need firmly know  WHAT  the conditional probability is.


            If  X  and  Y are the events in the universal set of events,  then  P(X|Y)  is  THIS  ratio


<pre>
                   P(X|Y) =  {{{P(X_and_Y)/P(Y)}}} .         
</pre>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Having this definition, &nbsp;let answer all questions, &nbsp;one after other.




<pre>
(a)  P(B|A) = {{{P(A_and_B)/P(A)}}} = {{{0.05/0.3}}} = {{{5/30}}} = {{{1/6}}}.    <U>ANSWER</U>


     Solved.    So, it is easy.




(b)  P(A|B').


     We should find  P(A &#8745; B').  The set  (A &#8745; B') are those elements/(events) of A that do not belong to B.

     From it, it is clear that the set (A &#8745; B') is  (A \ (A &#8745; B)).  

     Therefore,  P(A &#8745; B') = P(A) - P(A &#8745; B) = 0.3 - 0.05 = 0.25.


     Next, P(B') = 1 - P(B) = 1 - 0.4 = 0.6.


     THEREFORE,  P(A|B') = {{{0.25/0.6}}} = {{{25/60}}} = {{{5/12}}}.      <U>ANSWER</U>




(c)  P(A'|B').


     We should find  P(A' &#8745; B').  The set  (A' &#8745; B') are those elements/(events) of the universal set that belong NEITHER A NOR B.

     From it, it is clear that the set (A' &#8745; B') is  the complement of (A U B).  

     P(A U B) = P(A) + P(B) - P(A &#8745; B) = 0.3 + 0.4 - 0.05 = 0.65.

     Therefore,  P(A' &#8745; B') = 1 - P(A U B) = 1 - 0.65 = 0.35.


     Next, P(B') = 1 - P(B) = 1 - 0.4 = 0.6.


     THEREFORE,  P(A'|B') = {{{0.35/0.6}}} = {{{35/60}}} = {{{7/12}}}.      <U>ANSWER</U>
</pre>

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Solved.


All questions are answered.