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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 CONSISTING of 3 games.
Find the probability that A wins all three games.
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This problem is, actually, a standard problem of this class, but expressed in slightly unusual terms,
which may create difficulties to you in understanding its meaning.
Therefore, I will re-formulate this problem EQUIVALENTLY in familiar terms.
+------------------------------------------------------------------------+
| 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 ? |
+------------------------------------------------------------------------+
SOLUTION
The probability to win the first of the 3 games is P(win 1st) = = .
The probability to win the second of the 3 games is P(win 2nd) = .
The probability to win the third of the 3 games is P(win 1st) = = .
The probability to win all 3 games is
P = P(win 1st)*P(win 2nd)*P(win 3rd) = = = . ANSWER
Solved.
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The solution by @ewatrrr is INCORRECT, and her interpretation of the problem is INCORRECT, too.
So, ignore her post for your safety.