Two individuals, A and B, are finalist for a chess championship. They will play a sequence of games, each of which can result in a win for A, a win for B, or a draw. Outcomes of successive games are independent with P(A wins) =.3, P(B wins)= .2, P(draw)=.5. Each time a player wins a game he earns one point and opponent none, no points for draw. First player to win five points wins championship.
What is the probability that it takes just five games to obtain a champion?
First find out probability that A wins 5 games (.3x.3x.3x.3x.3=.0024)
Second find P(B wins) 5 games (.2x.2x.2x.2x.2=.00032)
Last add the probabilities of A and B (.00243+.00032=.00275)
Is this correct?
Yes, if you math is correct.
No, your calculation for the probability that it takes just five games to obtain a champion is not correct. To determine the probability that the championship is won in exactly five games, you need to consider all possible sequences of wins and draws that result in one player reaching five points.
Let's break it down step-by-step:
1. To win the championship in exactly five games, one player needs to win four games and the fifth game needs to result in a win or draw. There are two possible scenarios:
a) A wins four games and the fifth game results in a win or draw.
b) B wins four games and the fifth game results in a win or draw.
2. Calculate the probability for each scenario:
a) P(A wins 4 games) x P(A wins or draws 1 game) = (0.3)^4 x (0.3 + 0.5) = 0.3^4 x 0.8 = 0.003672.
b) P(B wins 4 games) x P(B wins or draws 1 game) = (0.2)^4 x (0.2 + 0.5) = 0.2^4 x 0.7 = 0.00112.
3. Add the probabilities of both scenarios to get the overall probability:
P(championship won in exactly five games) = 0.003672 + 0.00112 = 0.004792.
Therefore, the correct probability that it takes just five games to obtain a champion is approximately 0.004792 or 0.4792%.
No, your solution is not correct. Let's break it down step by step to find the correct probability.
To determine the probability that it takes just five games to obtain a champion, we need to consider all possible ways in which a player can win five points out of the first five games.
There are three possible outcomes for each game: A wins, B wins, or a draw. Let's calculate the probability of each outcome.
P(A wins) = 0.3
P(B wins) = 0.2
P(draw) = 0.5
To calculate the probability that A wins 5 out of 5 games, we need to consider the following scenarios:
1. A wins all 5 games (A-A-A-A-A): P(A wins) * P(A wins) * P(A wins) * P(A wins) * P(A wins) = (0.3) * (0.3) * (0.3) * (0.3) * (0.3) = 0.3^5
2. B wins 4 games, A wins 1 game (B-B-B-B-A): P(B wins) * P(B wins) * P(B wins) * P(B wins) * P(A wins) = (0.2) * (0.2) * (0.2) * (0.2) * (0.3) = 0.2^4 * 0.3
3. B wins 3 games, A wins 2 games (B-B-B-A-A): P(B wins) * P(B wins) * P(B wins) * P(A wins) * P(A wins) = (0.2) * (0.2) * (0.2) * (0.3) * (0.3) = 0.2^3 * 0.3^2
Now, we add up the probabilities of these scenarios:
P(five games to obtain a champion) = P(A wins 5 games) + P(B wins 4 games, A wins 1 game) + P(B wins 3 games, A wins 2 games)
= 0.3^5 + 0.2^4 * 0.3 + 0.2^3 * 0.3^2
Calculating this expression gives us the correct probability that it takes just five games to obtain a champion.