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Conditional probability is a fundamental concept in probability theory that describes the probability of an event occurring given that another event has already occurred. In this exercise, we are given a conditional probability density function (pdf) for random variable \(X_1\) given \(X_2 = x_2\). The function is expressed as:\[ f_{1 \mid 2}(x_{1} \mid x_{2}) = c_{1} \frac{x_{1}}{x_{2}^{2}} \]To determine the constant \(c_{1}\), we use the principle of normalization, which states that the total probability over all possible values should sum up to 1. Therefore, we integrate the conditional pdf from 0 to \(x_{2}\) with respect to \(x_{1}\):\[ \int_{0}^{x_{2}} c_{1} \frac{x_{1}}{x_{2}^{2}} \, dx_{1} = 1 \]Solving this integral gives us \(c_{1} = 2\). This means the function \(f_{1 \mid 2}(x_{1} \mid x_{2}) = \frac{2x_{1}}{x_{2}^{2}}\) when normalized for any given \(x_{2}\). Conditional probability helps us understand how the behavior of \(X_1\) is affected by knowing the value of \(X_2\).

Joint probability involves understanding the likelihood of two events occurring together, captured through a joint probability density function when dealing with continuous variables. In this scenario, we are tasked with finding the joint pdf of \(X_1\) and \(X_2\). The joint pdf can be determined by multiplying the conditional pdf of \(X_1\) given \(X_2\) by the marginal pdf of \(X_2\).Given:- Conditional pdf: \(f_{1 \mid 2}(x_{1} \mid x_{2}) = \frac{2x_{1}}{x_{2}^{2}}\) with \(c_{1} = 2\)- Marginal pdf: \(f_{2}(x_{2}) = 5x_{2}^{4}\) with \(c_{2} = 5\)The joint pdf is then computed as:\[ f(x_{1}, x_{2}) = f_{1 \mid 2}(x_{1} \mid x_{2}) \cdot f_{2}(x_{2}) = \frac{2 x_{1}}{x_{2}^{2}} \cdot 5 x_{2}^{4} = 10 x_{1} x_{2}^{2} \]where the product incorporates how \(X_1\) and \(X_2\) relate to each other. Joint probability is crucial for understanding how variables may be dependent on one another in probabilistic models.

Normalization in probability ensures that probability density functions (pdf) integrate to 1 over their entire range, establishing that the total probability is accounted for across all potential outcomes. Consider this exercise, where normalization is employed to solve for constants \(c_{1}\) and \(c_{2}\).### Finding \(c_{1}\)Normalization for the conditional pdf \(f_{1 \mid 2}(x_{1}|x_{2})\) requires:- Integrate \(f_{1 \mid 2}(x_{1}|x_{2})\) over \(x_{1}\) from 0 to \(x_{2}\).- Calculate: \(\int_{0}^{x_{2}} c_{1} \frac{x_{1}}{x_{2}^{2}} \, dx_{1} = 1\), leading to \(c_{1} = 2\).### Finding \(c_{2}\)Normalization for the marginal pdf \(f_{2}(x_{2})\) involves:- Integrate \(f_{2}(x_{2})\) over \(x_{2}\) from 0 to 1.- Compute: \(\int_{0}^{1} c_{2} x_{2}^{4} \, dx_{2} = 1\), giving \(c_{2} = 5\).By normalizing, we confirm that the calculated pdfs are valid probability distributions, offering a complete model for probability calculations.

Integration plays a pivotal role in probability, as it is used to calculate probabilities when dealing with continuous distributions. Integrals allow us to sum up infinite small probabilities across a range, providing likelihoods for specific events.### Conditional Probability ComputationTo compute \(P(\frac{1}{4} < X_{1} < \frac{1}{2} \mid X_{2} = \frac{5}{8})\):- Evaluate the integral of the joint pdf \(f(x_{1}, x_{2})\) with fixed \(x_{2} = \frac{5}{8}\) over the range of \(x_{1}\) from \(1/4\) to \(1/2\).- \(\int_{\frac{1}{4}}^{\frac{1}{2}} 10 x_{2}^{3} \, dx_{1} = 0.3125\) with substitution for \(x_{2}\).### Marginal Probability ComputationTo get \(P(\frac{1}{4} < X_{1} < \frac{1}{2})\):- Perform a double integral, first over \(x_{1}\) from \(1/4\) to \(1/2\) and then \(x_{2}\) from 0 to 1.- Solving \(\int_{0}^{1} \int_{\frac{1}{4}}^{\frac{1}{2}} 10 x_{2}^{3} \, dx_{1} \, dx_{2} = 0.015625\) gives the probability without the condition.Integration is critical, as it bridges the gap between probability calculations and continuous distributions, empowering the evaluation of outcomes over specified intervals.