Final answer:
To calculate the probability that both Alex and Bob will win at least 2 games, we need to determine the individual probabilities of each player winning and then combine them. We find that the probability of Bob winning a game is 1/4 and the probability of Alex winning a game is 3/4. By calculating the probabilities of different combinations where they both win at least 2 games and adding them up, we find that the probability is approximately 3.96%, which is closest to answer choice D (5%).
Step-by-step explanation:
In order to calculate the probability that both Alex and Bob will win at least 2 games, we need to determine the individual probabilities of each player winning and then combine them. Let's start by finding the probability of each player winning a game:
Let's assume that the probability of Bob winning a game is represented by 'x'. Since Alex is 3 times more likely to win, the probability of Alex winning a game is '3x'.
Now, we can set up the equation 'x + 3x = 1' because the sum of the probabilities of winning for both players must equal 1.
Simplifying the equation, we get '4x = 1'. Solving for 'x', we find that 'x = 1/4'.
So, the probability of Bob winning a game is '1/4' and the probability of Alex winning a game is '3/4'.
Now, let's calculate the probability that both of them will win at least 2 games:
To compute this, we need to find the probability of each combination where they both win at least 2 games and add them up.
P(Bob wins 2 games and Alex wins 2 games) = (1/4)^2 * (3/4)^2 = 9/256
P(Bob wins 2 games and Alex wins 3 games) = (1/4)^2 * (3/4)^3 = 27/1024
P(Bob wins 3 games and Alex wins 2 games) = (1/4)^3 * (3/4)^2 = 27/1024
P(Bob wins 3 games and Alex wins 3 games) = (1/4)^3 * (3/4)^3 = 81/4096
Adding up these probabilities, we get (9/256) + (27/1024) + (27/1024) + (81/4096) = 162/4096 = 0.0396.
Therefore, the probability that both Alex and Bob will win at least 2 games is approximately 3.96%, which is closest to answer choice D (5%).