Question 1177054: A and B plays 12 games of chess of which 6 are won by A, 4 are won by B and two ends on a tie. They agree to play a tournament consistian of 3 games. Find the probability that A wins all three games.
Found 2 solutions by ewatrrr, ikleyn:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
12 games: A wins 6, B wins 4 and 2T
play a tournament consisting of 3 games.
P(A wins 3 games) = (1/2)(1/2)(1/2) = 1/8
Wish You the Best in your Studies.
Answer by ikleyn(52778)  (Show Source):
You can put this solution on YOUR website! .
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.
~~~~~~~~~~~~~~~
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.
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