1. **State the problem:** A fair coin is tossed four times, and we want to find: a) The sample space. b) The probability that all four tosses are tails. c) The probability of getting exactly three heads. d) The probability of getting three heads in succession. 2. **Sample space:** Each toss has 2 possible outcomes (Head (H) or Tail (T)). For 4 tosses, the total number of outcomes is $2^4 = 16$. The sample space $S$ consists of all sequences of length 4 with H or T, e.g., $\{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT, TTHT\}$. 3. **Probability that all four tosses are tails:** Only one outcome is all tails: $TTTT$. Probability $P(\text{all tails}) = \frac{1}{16}$. 4. **Probability of getting exactly three heads:** Number of ways to choose which 3 tosses are heads out of 4 is $\binom{4}{3} = 4$. Each outcome has probability $\frac{1}{16}$. So, $P(\text{3 heads}) = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$. 5. **Probability of getting three heads in succession:** Possible sequences with 3 heads in a row are: - $HHHT$ - $THHH$ - $THHH$ (already counted) - $HHTH$ (no 3 in a row) Actually, the sequences with exactly 3 heads in a row are $HHHT$, $THHH$, and $THHH$ is repeated, so only two unique sequences. Also consider $THHH$ and $HTHH$? $HTHH$ has heads at positions 2,3,4 which is 3 in a row. So sequences with 3 heads in succession are: - $HHHT$ (positions 1-3) - $THHH$ (positions 2-4) - $HTHH$ (positions 2-4) Wait, $HTHH$ is heads at positions 2,3,4, so yes. So total sequences with 3 heads in a row: $HHHT$, $THHH$, $HTHH$. Probability $= \frac{3}{16}$. **Final answers:** - a) Sample space size: 16 sequences. - b) $P(\text{all tails}) = \frac{1}{16}$. - c) $P(\text{3 heads}) = \frac{1}{4}$. - d) $P(\text{3 heads in succession}) = \frac{3}{16}$.