91Ó°ÊÓ

In the context of probability distributions, a random variable is a numerical description of the outcome of a random phenomenon. In simpler terms, it is a variable that takes on different values based on the outcomes of a random event.

For instance, when you throw a die twice as in our exercise, each throw produces some outcome. Define these outcomes as \(X_1\) and \(X_2\). In our problem, the interest is on the minimum value between these two throws, denoted by \(X = \min(X_1, X_2)\). Thus, this makes \(X\) a random variable because it is determined based on the outcomes of the dice throws, which happen randomly.

Remember that random variables can be classified as either discrete or continuous. In this problem, \(X\) is a discrete random variable because it can only take on specific, countable values like 1, 2, 3, etc.

The sample space is a crucial concept in probability and refers to the set of all possible outcomes of a random experiment.

Consider a single die throw, which can result in any one of the six numbers from 1 to 6. Thus, the sample space for one die thrown is \( \{1, 2, 3, 4, 5, 6\} \). Now, when you roll two dice, each die independently contributes its own set of possibilities.

This results in a combined sample space that includes all possible ordered pairs of outcomes from the two dice. Therefore, the complete sample space when throwing two dice can be depicted as:

• \(\{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), \)
• \((2,1), (2,2), (2,3), (2,4), (2,5), (2,6), \)
• \(\ldots \)
• \((6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \}\)

There are 36 possible outcomes in total, as each die contributes 6 outcomes. Understanding the sample space is vital as it helps in determining the probability of events.

Probability calculation involves determining how likely certain events are to occur within the defined sample space. This requires understanding the number of favorable outcomes for the event and dividing it by the total possible outcomes.

In our specific exercise, we calculate probabilities for the random variable \(X = \min(X_1, X_2)\). For example, to find \(P(X = 1)\), you look for all pairs \((X_1, X_2)\) where at least one die shows a 1.

• The favorable outcomes are 11, with pairs like \((1,1), (1,2), (2,1), \ldots \).
• Thus, \(P(X=1) = \frac{11}{36}\)

The calculation process is repeated for each potential value of \(X\), considering at each step the different conditions that make \(X\) take a certain value. This division of favorable outcomes by total outcomes (36) gives the probability for each outcome, forming the basis for the probability distribution of \(X\).

A discrete probability distribution represents probabilities for a discrete random variable. Discrete means the variable can take on separate, distinct values like 0, 1, 2, etc. In our example, \(X\) is a discrete random variable that takes values from 1 to 6.

This probability distribution is described by a function that associates each possible value of \(X\) with its probability. The sum of all these probabilities equals 1, as it must cover all possible outcomes.

• For example, \(P(X = 1) = \frac{11}{36}\), \(P(X = 2) = \frac{9}{36}\), and so on.
• The probabilities are determined by the condition set for \(X\), like having a minimum outcome of 1 or 2 on the dice.

This distribution not only helps in understanding the likelihood of each outcome occurring but also assists in making predictions and decisions based on these probabilities. Discrete distributions, like the one we've outlined, form the cornerstone of probability theory, particularly in situations involving countable, distinct events.