Question 1143226
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Box A contains nine cards numbered 1 through 9,and box B contains five cards numbered 1 through 5. 
A box is chosen at random and a card is drawn. If the number is even find the probability that the card came from Box A.
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            The solution by the other post answers the question:

<pre>
                What is the probability to get an even number from box A ?
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;But the question in the problem is &nbsp;TOTALLY &nbsp;DIFFERENT, &nbsp;so that solution is &nbsp;IRRELEVANT.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Below find the correct solution.



<pre>
By the definition, the conditional probability is

    P(A) = P(even,A)/P(even,A+B),

where  P(A)        is the probability under the question that the number came from box A provided that it is even;

       P(even,A)   is the probability to get even number from box A;

       P(even,A+B) is the probability to get even number from either box A or box B.


Box A contains 9 numbers; of them, 4 numbers are even.
Box B contains 5 numbers; of them, 2 numbers are even.


Therefore, we have

    P(even,A+B) = {{{(1/2)*(4/9 + 2/5)}}} = {{{(1/2)*(20/45 + 18/45)}}} = {{{(1/2)*(38/45)}}};

    P(even,A)   = {{{(1/2)*(4/9)}}}.


Therefore,

    P(A) = {{{((4/9))/((38/45))}}} = {{{(2*5)/19}}} = {{{10/19}}} = 0.5263 = 52.63%  (approximately).     <U>ANSWER</U>
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