Question:

**Step 1: Identify the possible sequences** Let $H$ denote a head and $T$ denote a tail. The probability of $H$ is $p = \frac{1}{3}$ and the probability of $T$ is $q = \frac{2}{3}$. The experiment stops with head if the sequence ends in $HH$. The valid sequences are those alternating until they end in $HH$. These sequences are $HH$, $THH$, $HTHH$, $THTHH$, and so on. **Step 2: Calculate the probability using infinite geometric series** The sequences ending in $HH$ can be split into two groups:\nGroup 1 (starting with $H$): $HH$, $HTHH$, $HTHTHH$, $\dots$ with probabilities $p^2$, $p^2(pq)$, $p^2(pq)^2$, $\dots$\nGroup 2 (starting with $T$): $THH$, $THTHH$, $THTHTHH$, $\dots$ with probabilities $qp^2$, $qp^2(pq)$, $qp^2(pq)^2$, $\dots$ **Step 3: Sum the geometric series** The probability of Group 1 is an infinite geometric series: $\frac{p^2}{1-pq}$.\nThe probability of Group 2 is: $\frac{qp^2}{1-pq}$.\nThus, the total probability is $\frac{p^2(1+q)}{1-pq}$. **Step 4: Substitute the values** Substitute $p = \frac{1}{3}$ and $q = \frac{2}{3}$ into the formula:\nProbability $= \frac{(1/3)^2 (1 + 2/3)}{1 - (1/3)(2/3)} = \frac{(1/9)(5/3)}{1 - 2/9} = \frac{5/27}{7/9} = \frac{5}{21}$.