91Ó°ÊÓ

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.

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.

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.