Team A and B play 13 games of chess of which 6 are won by A, 4 are won by B and 3 end in a draw. They agree to play a tournament of 4 games. Find the probability that A wins all 4.
The probability that A wins a single game in the tournament is $\dfrac{6}{13}$. Since the outcomes of the games are independent, the probability that A wins all 4 games is $\left(\dfrac{6}{13}\right)^4 = \boxed{\dfrac{1296}{28561}}$.
To find the probability that A wins all 4 games in the tournament, we need to consider the total number of possible outcomes for the 4 games.
In each game, there are three possible outcomes: A wins, B wins, or the game ends in a draw.
Since there are 13 total games played between Team A and Team B, we know that all 13 games must be accounted for in these 3 possible outcomes.
From the information given, we know that A has won 6 games, B has won 4 games, and 3 games have ended in a draw. This means that out of the 13 games played, A has already won 6 games, B has already won 4 games, and there are 3 games remaining in the tournament.
To find the total number of possible outcomes for the remaining 3 games, we multiply the number of outcomes for each of the 3 games together: 3 * 3 * 3 = 27.
Out of these 27 possible outcomes, we need to find the number of outcomes where A wins all 3 games. Since we are interested in the scenario where A wins every game, we know that the outcome for each of the 3 games must be A wins. Therefore, there is only 1 favorable outcome where A wins all 3 games.
Therefore, the probability that A wins all 3 games in the tournament is 1/27.
To find the probability that A wins all 4 games, we need to know the total number of possible outcomes for the 4 games.
Since there are two possible outcomes for each game (A wins or B wins), the total number of possible outcomes for the 4 games is 2^4 = 16.
Now, we have the information about the previous 13 games they played. Out of those, 6 were won by A and 4 were won by B. This means that 13 games - 6 wins by A - 4 wins by B = 3 games ended in a draw.
Since a draw is another possible outcome, we need to account for it in our probability calculation. So, out of the 16 total possible outcomes, 3 outcomes are draws, leaving us with 16 - 3 = 13 outcomes where a winner is determined in the 4 games.
As we want to find the probability that A wins all 4 games, there is only one outcome where A wins all 4. Therefore, the probability that A wins all 4 games is 1/13, or approximately 0.0769.