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 short answer would be specific numerical value which you will get after addition of calculated probabilities from step 4.

Step-by-step solution

Identifying Possible Events

First, identify the possible events. There are three combinations where two variables are less than zero: event A where \(X_{1}\) and \(X_{2}\) are both less than zero, event B where \(X_{1}\) and \(X_{3}\) are less than zero, and event C where \(X_{2}\) and \(X_{3}\) are less than zero. Each event has corresponding \(X\) which will be greater than zero.

Computing the Probability for Each Event

The probability that a standard normal random variable \(X\) is less than zero is \(P(X<0) = 0.5\) because of the symmetry of the normal distribution, and the probability that \(X\) is greater than zero is the same, i.e., \(P(X > 0) = 0.5\). For \(N(2,4)\), refer to Z-table to compute \(P(X < 0)\) and \(P(X >0)\). Likewise, for \(N(-1, 1)\), refer to Z-table to compute \(P(X<0)\) and \(P(X>0)\).

Calculating the Individual Probabilities

Calculate the probability for event A which is \(P(X_{1}<0) \cdot P(X_{2}<0) \cdot P(X_{3}>0)\). Likewise, calculate for event B and event C.

Addition of Probabilities

As final step, add up all the probabilities calculated for event A, B, and C to obtain the total probability that exactly two out of three random variables are less than zero.