Question 1201644
.
Find the probability that the sum is as stated when a pair of dice is rolled. (Enter your answers as fractions.)
A. even and doubles
B. even or doubles
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



        The answers and the logic in the post by @mananth both are incorrect for both  (a)  and  (b).


        I came to bring correct answers with explanations.



<pre>
(a) even AND doubles.


    The condition says to count the pairs that are doubles and have even sum.

    This intersection of two conditions consists of 6 outcomes

        (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6).


    Notice that under this condition, each double pair has an even sum.


    So, in case (a)  the probability is  6/36 = 1/6.    <<<---===  <U>ANSWER</U>


    
(b) even or doubles


    The number of outcomes with even sum is 36/2 = 18.
    It is clear from symmetry.

    All doubles have even sum, so they fall into the even category.


    Thus, in case (b), the probability is  18/36 = 1/2.    <<<---===  <U>ANSWER</U>
</pre>

Solved correctly.



==========================

<H3>A notice to the visitor / to the creator of this problem</H3>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The problem formulation is provocatively incorrect.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;A correct formulation should be like THIS


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Find the probability that the conditions are satisfied when a pair of dice is rolled. 



Professional Math writers NEVER create provocative formulations, because their goal is not to confuse, but, in opposite - to teach.


Provocative formulations are the product of those unprofessional writers, whose goal is to confuse, but not to teach.


Do you feel the difference ? ?