Probability Theory – Problem 11: find the conditional probability of observing at least three tails;

A fair coin is tossed repeatedly until exactly two heads have appeared. Given that at least two tails were observed: (1) find the conditional probability of observing at least three tails; (2) find the conditional probability that the first toss resulted in heads.

Step 1. Define the basic events and probabilities. * When the experiment terminates, the last toss is necessarily heads, and among the first $N-1$ tosses there is exactly one head. * Let $A$ denote the event "at least two tails are observed." This means the number of tails $X \ge 2$, so the total number of tosses $N = X + 2 \ge 4$. * Since $P(A) = 1 - P(A^c)$, we first compute the probability of $A^c$ (fewer than two tails).

Step 2. Compute the probability of the complementary event $A^c$. * Case 1: Zero tails ($X=0$). The only possible sequence is $HH$. Probability: $P(X=0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. * Case 2: Exactly one tail ($X=1$). Total number of tosses $N=3$. The last toss is heads, and the first two tosses consist of one head and one tail. The possible sequences are: $HTH$, $THH$. Probability: $P(X=1) = 2 \times (\frac{1}{2})^3 = \frac{2}{8} = \frac{1}{4}$.

Step 3. Compute $P(A)$. * $P(A) = 1 - [P(X=0) + P(X=1)]$ * $P(A) = 1 - (\frac{1}{4} + \frac{1}{4}) = \frac{1}{2}$.

Step 4. Define the events and objective for part (1). * Let $B$ denote the event "at least three tails are observed," i.e., $X \ge 3$. * We need to compute $P(B|A) = \frac{P(AB)}{P(A)}$. * Since "at least three tails" implies "at least two tails," we have $B \subset A$, and therefore $P(AB) = P(B)$.

Step 5. Compute $P(B)$. * Using the previously computed $P(A)$, or equivalently $1 - P(X < 3)$. * We know $P(X \ge 2) = \frac{1}{2}$. * The event $X=2$ (exactly two tails) corresponds to sequences of length 4, where the last toss is heads and the first three tosses contain exactly one head and two tails. * The number of such sequences is $\binom{3}{1} = 3$ (namely $HTTH, THTH, TTHH$). * $P(X=2) = 3 \times (\frac{1}{2})^4 = \frac{3}{16}$. * $P(B) = P(X \ge 3) = P(X \ge 2) - P(X=2) = \frac{1}{2} - \frac{3}{16} = \frac{5}{16}$.

Step 6. Compute the conditional probability for part (1). * $P(B|A) = \frac{P(B)}{P(A)} = \frac{5/16}{1/2} = \frac{5}{16} \times 2 = \frac{5}{8}$.

Step 7. Define the events and objective for part (2). * Let $C$ denote the event "the first toss is heads." * We need to compute $P(C|A) = \frac{P(AC)}{P(A)}$. * The event $AC$ means: the first toss is heads, and exactly two heads appear in total (the second head causes the experiment to stop), and at least two tails are observed in between.

Step 8. Analyze the structure of the event $AC$. * Since the first toss is heads and the second head does not appear until the final toss, all intermediate tosses must be tails. * The sequence must therefore have the form: $H$ (first toss), $T, T, \dots, T$ ($k$ tails in the middle), $H$ (final toss). * We denote this as $H T^k H$. * The condition "at least two tails" requires $k \ge 2$.

Step 9. Compute $P(AC)$. * This is an infinite series of probabilities, where each qualifying sequence has a unique form. * For $k=2$ (2 tails): $HTTH$, with probability $(\frac{1}{2})^4 = \frac{1}{16}$. * For $k=3$ (3 tails): $HTTTH$, with probability $(\frac{1}{2})^5 = \frac{1}{32}$. * ... * $P(AC) = \sum_{k=2}^{\infty} (\frac{1}{2})^{k+2} = \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \dots$ * This is a geometric series with first term $a = \frac{1}{16}$ and common ratio $q = \frac{1}{2}$. * $P(AC) = \frac{1/16}{1 - 1/2} = \frac{1/16}{1/2} = \frac{1}{8}$.

Step 10. Compute the conditional probability for part (2). * $P(C|A) = \frac{P(AC)}{P(A)} = \frac{1/8}{1/2} = \frac{1}{4}$.

(1) Given that at least two tails were observed, the conditional probability of observing at least three tails is $\frac{5}{8}$. (2) Given that at least two tails were observed, the conditional probability that the first toss resulted in heads is $\frac{1}{4}$.