Question

Three coins are dropped on a table.

Answer 1

Answer:

(a) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

(b) $\frac{1}{8}$

(c) $\binom{3}{2} (\frac{1}{2})^3 = \frac{3}{8}$

(d) $(\frac{1}{2})^3 + \binom{3}{1} (\frac{1}{2})^3 = \frac{1}{8} + \frac{3}{8} = \frac{1}{2}$

(e) $1 - \frac{1}{2} = \frac{1}{2}$

(f) Yes, because $P(A \cap B) = 0$. Therefore $P(A \cup B) = P(A) + P(B) = \frac{3}{8} + \frac{1}{2} = \frac{7}{8}$

(g) No, because $P(A \cap C) = \frac{3}{8}$. Therefore $P(A \cup C) = P(A) + P(C) - P(A \cap C) = \frac{3}{8} + \frac{1}{2} - \frac{3}{8} = \frac{1}{2}$