Question

Let the mutually independent random variables \(X_{1}, X_{2}\), and \(X_{3}\) be \(N(0,1)\), \(N(2,4)\), and \(N(-1,1)\), respectively. Compute the probability that exactly two of these three variables are less than zero.

Short answer

The probability that exactly two of these three variables are less than zero is the sum of the three calculated probabilities from Step 3.

Step-by-step solution

Calculate Individual Probabilities

Calculate the probability that each random variable is less than zero:1. The random variable \(X_{1}\) is normally distributed with mean 0 and variance 1, i.e., \(X_{1} \sim N(0,1)\); thus it is a standard normal distribution. The probability that \(X_{1} < 0\) is 0.5.2. The random variable \(X_{2}\) is normally distributed with mean 2 and variance 4, i.e., \(X_{2} \sim N(2,4)\). We need to standardize it to become a standard normal distribution: \(Z_{2} = (X_{2} - 2) / 2\) and then find the probability that \(Z_{2} < (0 - 2) / 2 = -1\). This probability is approximately 0.1587.3. The random variable \(X_{3}\) is normally distributed with mean -1 and variance 1, i.e., \(X_{3} \sim N(-1,1)\). Thus it is a shifted standard normal distribution. The probability that \(X_{3} < 0\) corresponds to finding the probability that Z is less than 1, where Z is a standard normal variable. This probability is 1 minus the probability that Z is less than -1, which is approximately 0.8413.

Calculate the Combinations

As we need exactly two variables to be less than zero, we consider the following three cases:1. \(X_{1} < 0\) and \(X_{2} < 0\) and \(X_{3} \geq 0\)2. \(X_{1} < 0\) and \(X_{3} < 0\) and \(X_{2} \geq 0\)3. \(X_{2} < 0\) and \(X_{3} < 0\) and \(X_{1} \geq 0\)We add up the probabilities of these cases to provide the overall answer.

Compute the Overall Probability

Now, calculate the overall probability using the calculated individual probabilities and combination of gatherings:1. The probability for the first case is \(0.5 * 0.1587 * (1 - 0.8413)\).2. The probability for the second case is \(0.5 * (1 - 0.1587) * 0.8413\).3. The probability for the third case is \((1 - 0.5) * 0.1587 * 0.8413\).Adding these up gives the overall probability of exactly two random variables being less than zero.