Michael is playing tournament chess against a computer program and the probability he wins against the program at any given game is 0.25. Find the probability that on the given day Nigel wins for the first time on the 4th game played.
prob loses #1 = .75
prob loses #2 = .75 (2 lost in a row = .5625)
prob loses #3 = .75 (3 lost in a row = .421875)
prob then win #4 = .25*.421875 = 0.10546875
Well, Michael's chance of winning any given game is 0.25. So, the probability of him losing any given game is 0.75.
For Nigel to win for the first time on the 4th game played, Michael must have lost the first three games.
The probability of Michael losing the first game is 0.75, the second game is also 0.75, and the third game is also 0.75.
Since these events are independent, we can multiply the probabilities together to find the probability of this sequence happening.
0.75 * 0.75 * 0.75 = 0.421875
So, the probability that Nigel wins for the first time on the 4th game played is approximately 0.421875, or about 42.19%.
To find the probability that Nigel wins for the first time on the fourth game played, we need to consider all the possible outcomes that lead to this situation.
Since the probability of Nigel winning any single game is 0.25, the probability of him losing any single game is 1 - 0.25 = 0.75.
Let's consider the possible scenarios that lead to Nigel winning on the fourth game:
1. Nigel loses the first three games and wins the fourth: (0.75) * (0.75) * (0.75) * (0.25) = 0.10547.
2. Nigel loses the first two games, wins the third, and wins the fourth: (0.75) * (0.75) * (0.25) * (0.25) = 0.03516.
3. Nigel loses the first game, wins the second, wins the third, and wins the fourth: (0.75) * (0.25) * (0.25) * (0.25) = 0.01172.
Adding up these probabilities, we get:
0.10547 + 0.03516 + 0.01172 ≈ 0.15235.
Therefore, the probability that Nigel wins for the first time on the fourth game played is approximately 0.15235, or 15.235%.
To find the probability that Nigel wins for the first time on the 4th game played, we need to calculate the probability of the following events occurring in sequence:
1. Nigel loses the first 3 games.
2. Nigel wins the 4th game.
Since the probability of Nigel winning against the computer program in any given game is 0.25, the probability of him losing is 1 - 0.25 = 0.75.
To calculate the probability of multiple independent events occurring in sequence, we multiply the individual probabilities together.
Probability of Nigel losing the first 3 games: (0.75)^3 = 0.421875
Probability of Nigel winning the 4th game: 0.25
Therefore, the probability that Nigel wins for the first time on the 4th game played is 0.421875 * 0.25 = 0.10546875, or approximately 10.55%.