Question 1177054
.
A and B plays 12 games of chess of which 6 are wonby A, 4 are won by B and two ends on a tie. 
They agree to play a tournament {{{highlight(cross(consistian))}}} <U>CONSISTING</U> of 3 games. 
Find the probability that A wins all three games.
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This problem is, &nbsp;actually, &nbsp;a standard problem of this class, &nbsp;but expressed in slightly unusual terms,
which may create difficulties to you in understanding its meaning.


Therefore, &nbsp;I will re-formulate this problem &nbsp;EQUIVALENTLY &nbsp;in familiar terms.


<pre>
    +------------------------------------------------------------------------+
    |   In a bag, there are 6 black balls, 4 white balls and 2 blue balls.   |
    |   You draw 3 of these balls randomly without replacement.              |  
    |   What is the probability to draw 3 black balls ?                      |
    +------------------------------------------------------------------------+
</pre>


<U>SOLUTION</U>


<pre>
The probability to win the first of the 3 games is  P(win 1st) = {{{6/12}}} = {{{1/2}}}.


The probability to win the second of the 3 games is  P(win 2nd) = {{{5/11}}}.


The probability to win the third of the 3 games is  P(win 1st) = {{{4/10}}} = {{{2/5}}}.


The probability to win all 3 games is  


    P = P(win 1st)*P(win 2nd)*P(win 3rd) = {{{(1/2)*(5/11)*(2/5)}}} = {{{(5/11)*(1/5)}}} = {{{1/11}}}.    <U>ANSWER</U>
</pre>

Solved.


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The solution by @ewatrrr is &nbsp;INCORRECT, &nbsp;and her interpretation of the problem is &nbsp;INCORRECT, &nbsp;too.


So, &nbsp;ignore her post for your safety.