Solution. (a). Each coin lands two ways, so \(|S| = 2^4 = 16\) equally likely outcomes: \begin {align*} S = \{&HHHH,\ HHHT,\ HHTH,\ HTHH,\ THHH,\ HHTT,\ HTHT,\ HTTH,\\ &THHT,\ THTH,\ TTHH,\ HTTT,\ THTT,\ TTHT,\ TTTH,\ TTTT\}. \end {align*}

\(A\) excludes only \(TTTT\), so \(|A| = 15\). \(B\) collects the outcomes with \(0\), \(1\) or \(2\) heads, so \(|B| = 1+4+6 = 11\). \(C\) collects those with at least two tails, so \(|C| = 6+4+1 = 11\).

It is worth noticing that \(B\) and \(C\) are the same event. With four coins, at most two heads means at least two tails – the two descriptions pick out exactly the same 11 outcomes.

(b). \[P(A) = \frac {15}{16},\qquad P(B) = P(C) = \frac {11}{16}.\] \(A\cup B = S\), because the only outcome outside \(A\) is \(TTTT\), and \(TTTT\) is in \(B\): \[P(A\cup B) = 1.\] \(A\cap B\) is the set of outcomes with one or two heads, \(4+6 = 10\) of them: \[P(A\cap B) = \frac {10}{16} = \frac {5}{8}.\] \(A^c = \{TTTT\}\), which already lies in \(C\), so \(A^c\cup C = C\): \[P(A^c\cup C) = \frac {11}{16}.\]

Solution. Apply the two-event addition rule twice. Treat \(A\cup B\) as a single event and add \(C\) to it: \[P((A\cup B)\cup C) = P(A\cup B) + P(C) - P((A\cup B)\cap C).\] The first term expands by the same rule, \(P(A\cup B) = P(A)+P(B)-P(A\cap B)\). For the last term, distribute the intersection over the union: \[(A\cup B)\cap C = (A\cap C)\cup (B\cap C),\] and apply the addition rule once more to that: \[P((A\cap C)\cup (B\cap C)) = P(A\cap C)+P(B\cap C)-P((A\cap C)\cap (B\cap C)).\] Now \((A\cap C)\cap (B\cap C) = A\cap B\cap C\). Substituting the two expansions, \begin {align*} P(A\cup B\cup C) &= [P(A)+P(B)-P(A\cap B)] + P(C)\\ &\quad - [P(A\cap C)+P(B\cap C)-P(A\cap B\cap C)]\\ &= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)\\ &\quad +P(A\cap B\cap C), \end {align*}

as required.

The pattern is worth remembering: adding the three single probabilities counts each pairwise overlap twice, so the pairwise terms are subtracted; that subtraction removes the triple overlap three times having added it three times, so it must be added back once.

Solution. “At random” from the square means the probability of landing in a region equals that region’s area, since the whole square has area 1.

It is easier to work with the losing event. The player loses when both coordinates are at most \(\beta \), that is when the point falls in the smaller square \(0\leq x\leq \beta \), \(0\leq y\leq \beta \), which has area \(\beta ^{2}\). Hence \[P(\text {win}) = 1 - \beta ^{2}.\] The game is fair when the player wins half the time: \[1-\beta ^{2} = \frac {1}{2} \implies \beta ^{2} = \frac {1}{2} \implies \beta = \frac {1}{\sqrt {2}} = \frac {\sqrt {2}}{2} \approx 0.707.\] (The negative root is discarded because \(\beta \) must lie in \([0,1]\).)

Solution. (a). Order does not matter, so an outcome is a pair \(\{a,b\}\) with \(a\neq b\). There are \(\binom {5}{2}=10\) of them: \[S = \{12,\,13,\,14,\,15,\,23,\,24,\,25,\,34,\,35,\,45\}.\]

(b). The pairs differing by 1 are \(12, 23, 34, 45\) – four of the ten. All outcomes are equally likely, so \[P(\text {differ by }1) = \frac {4}{10} = \frac {2}{5}.\]

Solution. There are 20 equally likely cards.

(a). The multiples of 4 are \(4, 8, 12, 16, 20\), so \(P = \frac {5}{20} = \frac {1}{4}\).

(b). The numbers greater than 15 are \(16, 17, 18, 19, 20\), so \(P = \frac {5}{20} = \frac {1}{4}\).

(c). Both conditions hold for 16 and 20 only, so \(P = \frac {2}{20} = \frac {1}{10}\).

For parts (d) and (e), the card already drawn is divisible by 4 and so is even. It is not replaced, which leaves 19 cards: 9 even and 10 odd.

(d). \(P(\text {second is even}) = \frac {9}{19}\).

(e). \(P(\text {second is odd}) = \frac {10}{19}\).

Solution. The die and the two coins are independent, and the sample space has \(6\times 2\times 2 = 24\) equally likely outcomes.

(a). Two heads has probability \(\tfrac 14\); a number less than 3 (that is, 1 or 2) has probability \(\tfrac 26=\tfrac 13\). Hence \[P = \frac {1}{4}\times \frac {1}{3} = \frac {1}{12}.\]

(b). The coins differ in 2 of the 4 ways, probability \(\tfrac 12\); the die is even with probability \(\tfrac 36=\tfrac 12\). Hence \(P = \tfrac 12\times \tfrac 12=\tfrac 14\).

(c). The die is odd with probability \(\tfrac 12\) and the coins match with probability \(\tfrac 12\), so \(P = \tfrac 14\).

(d). The die shows 6 with probability \(\tfrac 16\); at least one head has probability \(1-\tfrac 14 = \tfrac 34\). Hence \[P = \frac {1}{6}\times \frac {3}{4} = \frac {1}{8}.\]

Solution. The tosses are independent with \(P(H)=\tfrac 23\) and \(P(T)=\tfrac 13\).

(a). No heads means three tails: \[P = \left (\frac {1}{3}\right )^{3} = \frac {1}{27}.\]

(b). With three tosses, more heads than tails means two heads or three heads: \begin {align*} P &= \binom {3}{2}\left (\frac {2}{3}\right )^{2}\left (\frac {1}{3}\right ) + \left (\frac {2}{3}\right )^{3}\\ &= 3\cdot \frac {4}{9}\cdot \frac {1}{3} + \frac {8}{27} = \frac {12}{27}+\frac {8}{27} = \frac {20}{27}. \end {align*}