Question

Let \(A_{1}=\\{(x, y): x \leq 2, y \leq 4\\}, A_{2}=\\{(x, y): x \leq 2, y \leq 1\\}, A_{3}=\) \(\\{(x, y): x \leq 0, y \leq 4\\}\), and \(A_{4}=\\{(x, y): x \leq 0 y \leq 1\\}\) be subsets of the space \(\mathcal{A}\) of two random variables \(X\) and \(Y\), which is the entire two-dimensional plane. If \(P\left(A_{1}\right)=\frac{7}{8}, P\left(A_{2}\right)=\frac{4}{8}, P\left(A_{3}\right)=\frac{3}{8}\), and \(P\left(A_{4}\right)=\frac{2}{8}\), find \(P\left(A_{5}\right)\), where \(A_{5}=\\{(x, y): 0

Short answer

Hence, the probability of \(A_{5}\) is \(\frac{1}{4}\)

Step-by-step solution

Identify A5 within A1

By recognizing that the region defined as A5 is actually within the region defined as A1, or \(A_{5} \subseteq A_{1}\). Because \(A_{1}\) includes all (x,y) where x is less than or equal to 2 and y is less than or equal to 4. That includes the region A5 where x is greater than 0 and less than or equal to 2, and y is greater than 1 and less than or equal to 4.

Subtract A2,A3,A4 from A1

To get to the exact region described as A5, we need to subtract regions A2, A3 and A4 from A1. Thus, the probability of A5 is given by \(P(A_{5}) = P(A_{1}) - P(A_{2}) - P(A_{3}) + P(A_{4})\). Explanation: Subtracting \(P(A_{2})\) and \(P(A_{3})\) from \(P(A_{1})\) removes the area of the two rectangles (A2 and A3) that are outside of A5. However, doing this also removes the region A4 twice. So, we have to add back \(P(A_{4})\) to correct this.

Substitute given probabilities

Substitute given probabilities into the equation from step 2. So, \(P(A_{5}) = \frac{7}{8} - \frac{4}{8} - \frac{3}{8} + \frac{2}{8} = \frac{2}{8} = \frac{1}{4}\).