Probability of Coin Tossing Experiment Ending with Consecutive Heads

To find the probability that the experiment stops with a head (i.e., ending with two consecutive heads, $HH$), we first define the probabilities for a single coin toss:

• Probability of getting a Head: $p = P(H) = \frac{1}{3}$
• Probability of getting a Tail: $q = P(T) = 1 - p = \frac{2}{3}$

The experiment continues tossing the coin until two consecutive outcomes are identical (either $HH$ or $TT$).

Method 1: Using Infinite Series

For the experiment to stop with two consecutive Heads, the sequence of outcomes must end with $HH$, and all preceding tosses must alternate in outcome so that no two adjacent tosses are identical before the final $HH$.

The possible winning sequences can be divided into two cases depending on the first toss:

1. Sequences starting with $H$:

   • Length 2: $HH \implies p^2$
   • Length 4: $HTHH \implies p \cdot q \cdot p \cdot p = p^3 q$
   • Length 6: $HTHTHH \implies p^4 q^2$
   • In general, for length $2k$ (where $k \ge 1$): $p^{k+1} q^{k-1}$

   Sum of probabilities for Case 1: $S_1 = p^2 + p^3 q + p^4 q^2 + \dots = \frac{p^2}{1 - pq}$

2. Sequences starting with $T$:

   • Length 3: $THH \implies q \cdot p \cdot p = q p^2$
   • Length 5: $THTHH \implies q^2 p^3$
   • Length 7: $THTHTHH \implies q^3 p^4$
   • In general, for length $2k+1$ (where $k \ge 1$): $q^k p^{k+1}$

   Sum of probabilities for Case 2: $S_2 = q p^2 + q^2 p^3 + q^3 p^4 + \dots = \frac{q p^2}{1 - pq}$

Adding both cases, the total probability $P$ that the experiment stops with $HH$ is: $P = S_1 + S_2 = \frac{p^2 + q p^2}{1 - pq} = \frac{p^2(1 + q)}{1 - pq}$

Substituting $p = \frac{1}{3}$ and $q = \frac{2}{3}$: $p^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$ $1 + q = 1 + \frac{2}{3} = \frac{5}{3}$ $1 - pq = 1 - \left(\frac{1}{3}\right)\left(\frac{2}{3}\right) = 1 - \frac{2}{9} = \frac{7}{9}$

$P = \frac{\frac{1}{9} \times \frac{5}{3}}{\frac{7}{9}} = \frac{\frac{5}{27}}{\frac{7}{9}} = \frac{5}{27} \times \frac{9}{7} = \frac{5}{21}$

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Method 2: Using Conditional Probability (Recurrence)

Let $P_H$ be the conditional probability that the experiment ends with $HH$ given that the previous single outcome was $H$. Let $P_T$ be the conditional probability that the experiment ends with $HH$ given that the previous single outcome was $T$.

Starting from the initial state (before any toss):

• The first toss is $H$ with probability $p$, moving the state to $H$.
• The first toss is $T$ with probability $q$, moving the state to $T$.

So, the required probability is: $P = p P_H + q P_T$

From state $H$:

• If the next toss is $H$ (probability $p$), the sequence $HH$ occurs, and the experiment stops with $HH$ (success, probability $1$).
• If the next toss is $T$ (probability $q$), the sequence is $HT$, which moves us to state $T$.

Thus: $P_H = p(1) + q P_T = p + q P_T \quad \text{--- (1)}$

From state $T$:

• If the next toss is $T$ (probability $q$), the sequence $TT$ occurs, and the experiment stops with $TT$ (failure, probability $0$).
• If the next toss is $H$ (probability $p$), the sequence is $TH$, which moves us to state $H$.

Thus: $P_T = q(0) + p P_H = p P_H \quad \text{--- (2)}$

Substituting equation (2) into (1): $P_H = p + q(p P_H) \implies P_H(1 - pq) = p \implies P_H = \frac{p}{1 - pq}$

Then, $P_T = \frac{p^2}{1 - pq}$

Substituting $P_H$ and $P_T$ back into $P$: $P = p \left(\frac{p}{1 - pq}\right) + q \left(\frac{p^2}{1 - pq}\right) = \frac{p^2(1 + q)}{1 - pq}$

Plugging in $p = \frac{1}{3}$ and $q = \frac{2}{3}$: $P = \frac{\left(\frac{1}{3}\right)^2 \left(1 + \frac{2}{3}\right)}{1 - \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)} = \frac{\frac{1}{9} \times \frac{5}{3}}{\frac{7}{9}} = \frac{5}{21}$

Therefore, the correct option is (B).