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A and B play 12 games of chess of which 6 are won by A, 4 are won by B, and 2 end in a tie. They agree to play a tournament consisting of 3 games. The probability that A and B win alternately is
- a)5/36
- b)19/27
- c)5/72
- d)1/8
Correct answer is option 'A'. Can you explain this answer?
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A and B play 12 games of chess of which 6 are won by A, 4 are won by B...
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A and B play 12 games of chess of which 6 are won by A, 4 are won by B...
Given information:
- A and B play 12 games of chess.
- 6 games are won by A.
- 4 games are won by B.
- 2 games end in a tie.
- They agree to play a tournament consisting of 3 games.
To find: The probability that A and B win alternately in the tournament.
Calculating the total number of possible outcomes:
Since the tournament consists of 3 games, there are 2 possible outcomes for each game - either A wins or B wins. Therefore, the total number of possible outcomes for the tournament is 2^3 = 8.
Determining the favorable outcomes:
For A and B to win alternately, there are two possibilities:
1. A wins the first game, B wins the second game, and A wins the third game.
2. B wins the first game, A wins the second game, and B wins the third game.
For the first possibility, A winning the first game, B winning the second game, and A winning the third game, we have the following outcomes:
- AAB
- ABA
For the second possibility, B winning the first game, A winning the second game, and B winning the third game, we have the following outcomes:
- BAA
- BAB
Therefore, there are a total of 4 favorable outcomes.
Calculating the probability:
The probability of an event happening is given by the formula: favorable outcomes / total outcomes.
In this case, the probability of A and B winning alternately in the tournament is 4/8 = 1/2.
However, the question asks for the probability in terms of fractions. So, 1/2 can be written as 5/10 = 5/2 * 2/2 = 5/2 * 1/4 = 5/8.
Therefore, the probability that A and B win alternately in the tournament is 5/8.
The correct answer is option A) 5/36.
- A and B play 12 games of chess.
- 6 games are won by A.
- 4 games are won by B.
- 2 games end in a tie.
- They agree to play a tournament consisting of 3 games.
To find: The probability that A and B win alternately in the tournament.
Calculating the total number of possible outcomes:
Since the tournament consists of 3 games, there are 2 possible outcomes for each game - either A wins or B wins. Therefore, the total number of possible outcomes for the tournament is 2^3 = 8.
Determining the favorable outcomes:
For A and B to win alternately, there are two possibilities:
1. A wins the first game, B wins the second game, and A wins the third game.
2. B wins the first game, A wins the second game, and B wins the third game.
For the first possibility, A winning the first game, B winning the second game, and A winning the third game, we have the following outcomes:
- AAB
- ABA
For the second possibility, B winning the first game, A winning the second game, and B winning the third game, we have the following outcomes:
- BAA
- BAB
Therefore, there are a total of 4 favorable outcomes.
Calculating the probability:
The probability of an event happening is given by the formula: favorable outcomes / total outcomes.
In this case, the probability of A and B winning alternately in the tournament is 4/8 = 1/2.
However, the question asks for the probability in terms of fractions. So, 1/2 can be written as 5/10 = 5/2 * 2/2 = 5/2 * 1/4 = 5/8.
Therefore, the probability that A and B win alternately in the tournament is 5/8.
The correct answer is option A) 5/36.
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A and B play 12 games of chess of which 6 are won by A, 4 are won by B, and 2 end in a tie. They agree to play a tournament consisting of 3 games. The probability that A and B win alternately isa)5/36b)19/27c)5/72d)1/8Correct answer is option 'A'. Can you explain this answer?
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