How to Solve Discrete Random Variables — Univariate RV — geometric distribution Questions on Exam P

A geometric count on \(1,2,\ldots\) has mean \(1/p=8\), so test positivity probability is \(p=1/8\) and failure probability is \(q=1-p=\frac{7}{8}\). Independence makes successive failure probabilities multiply. First positive on test \(6\) requires five failures then success: \(q^5p=(\frac{7}{8})^5(1/8)=0.0641\). The evaluated result is \(0.0641\), which matches the correct option. After simplifying the displayed expression, the resulting value is \(0.0641\). This is the numerical value of the probability the first positive occurs on test 6.