How to Solve Discrete Random Variables — Univariate RV — negative binomial (k-th success) Questions on Exam P

The third head occurs on toss \(5\) if toss \(5\) is a head and exactly 2 heads occur in the first \(4\) tosses. This probability is \(\Pr(\text{stop at }5)=\binom{4}{2}\left(\dfrac{1}{2}\right)^{4}\left(\dfrac{1}{2}\right)=\dfrac{\binom{4}{2}}{2^{5}}\). If toss \(4\) is also a head, then toss \(5\) and toss \(4\) are heads, requiring exactly 1 head among the first \(3\) tosses. This joint probability is \(\binom{3}{1}\left(\dfrac{1}{2}\right)^{5}=\dfrac{5-2}{2^{5}}\). After simplifying the displayed expression, the resulting value is \(0.0938\). This is the numerical value of \(\Pr(\text{the third head occurs on toss }5\text{ and toss }4\text{ is a head})\).