91Ó°ÊÓ

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.

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.

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.

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.