Probability Theory – Problem 63: Compute the conditional probability of tossing at least three tails;

A fair coin is tossed repeatedly until two heads are obtained. Given that at least two tails were tossed: (1) Compute the conditional probability of tossing at least three tails; (2) Compute the conditional probability that the first toss was a head.

1. Define the basic events and probabilities. * The experiment stops when the last toss is a head, and among the first $N-1$ tosses there is exactly one head. * Let $A$ be the event "at least two tails were tossed." 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: $HH$. 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 a head, and among the first two tosses there is one head and one tail. 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" necessarily implies "at least two tails," i.e., $B \subset A$, we have $P(AB) = P(B)$.

2. Compute $P(B)$. * Using the already computed $P(A)$ for a recursive approach, or using $1 - P(X < 3)$. * 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 a head and exactly one head and two tails 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 a head." * We need to compute $P(C|A) = \frac{P(AC)}{P(A)}$. * The event $AC$ means: the first toss is a head, and exactly two heads appear in total (the second head causes the process to stop), and there are at least two tails in between.

2. Analyze the structure of the event $AC$. * Since the first toss is a head and the second head only appears at the very end (stopping the process), 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 sum of probabilities over an infinite series, where each qualifying sequence has a unique form. * For $k=2$ (2 tails): $HTTH$, probability $(\frac{1}{2})^4 = \frac{1}{16}$. * For $k=3$ (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 was a head is $\frac{1}{4}$.