Probability Theory – Problem 67: Compute the conditional probability of obtaining at least three tails;

A fair coin is tossed repeatedly until two heads have appeared. Given that at least two tails have occurred: (1) Compute the conditional probability of obtaining at least three tails; (2) Compute the conditional probability that the first toss is heads.

1. Define the basic events and probabilities. * The experiment stops when the last toss is heads and exactly two heads have appeared in total among the first $N$ tosses. * Let $A$ be the event "at least two tails occurred." 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 $P(A^c)$ (fewer than two tails).

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

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

1. Set up the events and objective. * Let $B$ be the event "at least three tails," 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," i.e., $B \subset A$, we have $P(AB) = P(B)$.

2. Compute $P(B)$. * Use the already computed $P(A)$ for a recursive approach, or use $1 - P(X < 3)$. * We know $P(X \ge 2) = \frac{1}{2}$. * The event $X=2$ (exactly two tails) corresponds to sequences of length 4, with the last toss being heads and exactly one heads among the first three tosses. * The number of such sequences is $\binom{3}{1} = 3$ (namely $HTTH, THTH, TTHH$). * $P(X=2) = 3 \times (\frac{1}{2})^4 = \frac{3}{16}$. * $P(B) = P(X \ge 3) = P(X \ge 2) - P(X=2) = \frac{1}{2} - \frac{3}{16} = \frac{5}{16}$.

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

1. Set up the events and objective. * Let $C$ be 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 heads causes the process to stop), and at least two tails occur in between.

2. Analyze the structure of the event $AC$. * Since the first toss is heads and the second heads only appears at the very end (causing the process to stop), all intermediate tosses must be tails. * The sequence must have the form: $H$ (first toss), $T, T, \dots, T$ ($k$ tails in the middle), $H$ (end). * Denote this as $H T^k H$. * The problem requires "at least two tails," i.e., $k \ge 2$.

3. Compute $P(AC)$. * This is a probability sum over an infinite series, where each qualifying sequence has a unique form. * $k=2$ (2 tails): $HTTH$, probability $(\frac{1}{2})^4 = \frac{1}{16}$. * $k=3$ (3 tails): $HTTTH$, 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 $q = \frac{1}{2}$. * $P(AC) = \frac{1/16}{1 - 1/2} = \frac{1/16}{1/2} = \frac{1}{8}$.

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

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