Question

Contest. You play two games against the same opponent. The probability you win the first game is \(0.4\). If you win the first game, the probability you also win the second is \(0.2\). If you lose the first game, the probability that you win the second is \(0.3\). a) Are the two games independent? Explain. b) What's the probability you lose both games? c) What's the probability you win both games? d) Let random variable \(X\) be the number of games you win. Find the probability model for \(X\). e) What are the expected value and standard deviation?

Short answer

a) No, they are not independent. b) 0.42 c) 0.08 d) X ~ {0:0.42, 1:0.5, 2:0.08} e) E[X]=0.66, σ(X)≈0.62.

Step-by-step solution

Check for Independence

Two events, A and B, are independent if the probability of A and B occurring together is the product of their probabilities, i.e., \[ P(A \cap B) = P(A) \cdot P(B) \]. In this case, the probability of winning the second game given we won the first one is not equal to the probability of simply winning the second game, so the games are not independent.

Calculate Probability of Losing Both Games

To find the probability of losing both games, first calculate the probability of losing the first game: \[ P(L_1) = 1 - P(W_1) = 1 - 0.4 = 0.6 \]. If you lose the first game, the probability you lose the second game is: \[ P(L_2 | L_1) = 1 - P(W_2 | L_1) = 1 - 0.3 = 0.7 \]. So, the probability of losing both games is:\[ P(L_1 \cap L_2) = P(L_1) \cdot P(L_2 | L_1) = 0.6 \cdot 0.7 = 0.42 \].

Calculate Probability of Winning Both Games

To calculate the probability of winning both games, the probability you win the first game:\[ P(W_1) = 0.4 \]. If you win the first game, the probability you win the second game is:\[ P(W_2 | W_1) = 0.2 \]. Thus, the probability of winning both games is:\[ P(W_1 \cap W_2) = P(W_1) \cdot P(W_2 | W_1) = 0.4 \cdot 0.2 = 0.08 \].

Define the Random Variable Model

Let X represent the number of games won: X = 0, 1, or 2. Calculate each probability:- \( P(X = 0) \): Probability of losing both games. \( P(X = 0) = P(L_1 \cap L_2) = 0.42 \).- \( P(X = 2) \): Probability of winning both games. \( P(X = 2) = P(W_1 \cap W_2) = 0.08 \).- \( P(X = 1) \): Win one game, calculated as one win and one loss: \[ P(X = 1) = P(W_1 \cap L_2) + P(L_1 \cap W_2) \] - \( P(W_1 \cap L_2) = P(W_1) \cdot P(L_2 | W_1) = 0.4 \cdot 0.8 = 0.32 \) - \( P(L_1 \cap W_2) = P(L_1) \cdot P(W_2 | L_1) = 0.6 \cdot 0.3 = 0.18 \) - \( P(X = 1) = 0.32 + 0.18 = 0.5 \).Thus, the probability model for X is:\[ X \sim\begin{cases}0 & \text{with probability } 0.42 \ 1 & \text{with probability } 0.5 \ 2 & \text{with probability } 0.08 \end{cases}\].

Calculate Expected Value and Standard Deviation

The expected value \( E[X] \) of a discrete random variable is calculated as:\[ E[X] = \sum x \cdot P(X = x) = 0 \cdot 0.42 + 1 \cdot 0.5 + 2 \cdot 0.08 = 0 + 0.5 + 0.16 = 0.66 \].The variance \( Var(X) \) is:\[ Var(X) = \sum (x - E[X])^2 \cdot P(X = x) = (0 - 0.66)^2 \cdot 0.42 + (1 - 0.66)^2 \cdot 0.5 + (2 - 0.66)^2 \cdot 0.08 \]\[ = 0.4356 \cdot 0.42 + 0.1156 \cdot 0.5 + 1.7764 \cdot 0.08 \]\[ = 0.183552 + 0.0578 + 0.142112 = 0.383464 \]. Thus, the standard deviation \( \sigma(X) \) is:\[ \sigma(X) = \sqrt{Var(X)} = \sqrt{0.383464} \approx 0.62 \].

Key concepts

Random Variable

In probability and statistics, a random variable is a variable that represents the outcomes of a random phenomenon. In our contest exercise, we define the random variable \( X \) as the number of games you win against your opponent. This can take the values 0, 1, or 2, depending on how many games you actually win. Each of these outcomes has a certain probability associated with it.
To fully describe a random variable, we create a probability model. For \( X \), the probability distribution is given as follows:

• \( X = 0 \) with a probability of 0.42 (both games are lost).
• \( X = 1 \) with a probability of 0.50 (you win one game and lose one).
• \( X = 2 \) with a probability of 0.08 (you win both games).

Understanding random variables is key to describing and analyzing probabilities in a formal, mathematical way.

Expected Value

The expected value of a random variable is a measure of its central tendency. It provides a single number that represents the average or mean outcome if the random process were to be repeated many times. In our contest exercise, we calculate the expected value \( E[X] \) of the random variable \( X \) using the formula: \[ E[X] = \sum x \cdot P(X = x) \]So for our example:

• Multiply each possible value of \( X \) by its probability.
• Add all these products together.

Thus,\[ E[X] = 0 \cdot 0.42 + 1 \cdot 0.5 + 2 \cdot 0.08 = 0 + 0.5 + 0.16 = 0.66 \]This means that, on average, you can expect to win about 0.66 games in this contest.

Standard Deviation

The standard deviation measures the amount of variation in a set of values. It gives us an idea of how spread out the values of a random variable might be. A higher standard deviation indicates that the values are more spread out over a range.To determine the standard deviation, we first calculate the variance, which is the average of the squared differences from the expected value. For the random variable \( X \), the formula is:\[ Var(X) = \sum (x - E[X])^2 \cdot P(X = x) \]In our exercise:\[ Var(X) = (0 - 0.66)^2 \cdot 0.42 + (1 - 0.66)^2 \cdot 0.5 + (2 - 0.66)^2 \cdot 0.08 \]Carrying out the calculations results in a variance of approximately 0.383, and the standard deviation \( \sigma(X) \) is the square root of the variance:\[ \sigma(X) = \sqrt{0.383} \approx 0.62 \]This indicates a moderate level of variability in the number of games you might win.

Independence of Events

In probability, two events are considered independent if the occurrence of one does not affect the probability of the other. Mathematically, this is expressed as \( P(A \cap B) = P(A) \cdot P(B) \).In our contest problem, we are determining if winning the two games is independent. To check for independence, observe that the probability of winning the second game depends on whether you have won the first. Specifically:

• Probability of winning the second game, given you won the first, \( P(W_2 | W_1) = 0.2 \).
• Probability of winning the second game, given you lost the first, \( P(W_2 | L_1) = 0.3 \).

Because these probabilities are different and depend on the outcome of the first game, the two games are not independent.
Understanding independence is crucial for analysis and can significantly simplify calculating probabilities when events are indeed independent.