Compound probability

Outcomes: HH, HT, TH, TT (4 outcomes).
$6 \times 6 = 36$ outcomes.
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{5}{8} = \dfrac{25}{64}$ .
$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .
Even numbers are $2$ and $4$ , so $P(\text{even}) = \dfrac{2}{4} = \dfrac{1}{2}$ . $P(\text{both even}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$ .
$P(\text{spade}) = \dfrac{13}{52}$ , $P(\text{ace}) = \dfrac{4}{52}$ , $P(\text{spade and ace}) = \dfrac{1}{52}$ . $P(\text{spade or ace}) = \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$ .
Relative frequency $= \dfrac{114}{200} = 0.57$ .
Yes, they are mutually exclusive. A single die cannot show $6$ (even) and an odd number at the same time — the events have no outcomes in common.
$2 \times 2 = 4$ outcomes.
$P \approx \dfrac{18}{50} = 0.36$ .

Total $= 12$ . $P(\text{1st blue}) = \dfrac{4}{12} = \dfrac{1}{3}$ . $P(\text{2nd blue} \mid \text{1st blue}) = \dfrac{3}{11}$ . $P(\text{both blue}) = \dfrac{1}{3} \times \dfrac{3}{11} = \dfrac{3}{33} = \dfrac{1}{11}$ .
$P(\text{both aces}) = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{12}{2652} = \dfrac{1}{221}$ .
(a) $P(\text{both defective}) = \dfrac{3}{20} \times \dfrac{2}{19} = \dfrac{6}{380} = \dfrac{3}{190}$ . (b) $P(\text{neither defective}) = \dfrac{17}{20} \times \dfrac{16}{19} = \dfrac{272}{380} = \dfrac{68}{95}$ . (c) $P(\text{exactly one}) = 1 - \dfrac{3}{190} - \dfrac{68}{95} = 1 - \dfrac{3}{190} - \dfrac{136}{190} = \dfrac{51}{190} = \dfrac{51}{190}$ .
$P(A \text{ and } B) = 0.3 \times 0.5 = 0.15$ . $P(A \text{ or } B) = 0.3 + 0.5 - 0.15 = 0.65$ .
$P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$ . $P(\text{heads}) = \dfrac{1}{2}$ . Events are independent, so $P(\text{even and heads}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$ .
$P(\text{sport or music}) = \dfrac{12 + 8 - 3}{30} = \dfrac{17}{30}$ .
Independent events: the occurrence of one does not affect the probability of the other. Example: rolling a die and flipping a coin — the die result does not change the coin probability. Mutually exclusive events: the events cannot both occur at the same time. Example: rolling a $3$ and rolling a $5$ on a single die. Note: mutually exclusive events with non-zero probabilities are never independent (if one occurs, the probability of the other becomes $0$ ).
Simulation steps: (i) Assign heads $=$ tails outcome for a coin. (ii) Flip $4$ coins and record the number of tails. (iii) If $3$ or $4$ tails, record a success. (iv) Repeat for $100$ or more trials. (v) Estimate $P(\text{at least 3 tails}) = \dfrac{\text{number of successes}}{\text{total trials}}$ . Theoretical value: $P = \dbinom{4}{3}\left(\dfrac{1}{2}\right)^4 + \dbinom{4}{4}\left(\dfrac{1}{2}\right)^4 = \dfrac{4+1}{16} = \dfrac{5}{16} = 0.3125$ .

Let $D =$ has disease, $T^+ =$ tests positive. $P(D) = 0.02$ , $P(T^+ \mid D) = 0.95$ , $P(T^+ \mid \text{no } D) = 0.10$ . By tree diagram: $P(T^+) = 0.02 \times 0.95 + 0.98 \times 0.10 = 0.019 + 0.098 = 0.117$ . $P(D \mid T^+) = \dfrac{0.019}{0.117} \approx 0.162$ . Only about $16\%$ of positive results are true positives — the low disease rate means most positives are false alarms.
$P(A \text{ and } B) = P(A) + P(B) - P(A \text{ or } B) = 0.4 + 0.5 - 0.7 = 0.2$ . If independent: $P(A) \times P(B) = 0.4 \times 0.5 = 0.2$ . Since $P(A \text{ and } B) = P(A) \times P(B)$ , yes, $A$ and $B$ are independent.
It is easier to find $P(\text{no red}) = P(\text{all white}) = \dfrac{4}{9} \times \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{24}{504} = \dfrac{1}{21}$ . So $P(\text{at least one red}) = 1 - \dfrac{1}{21} = \dfrac{20}{21}$ .
$P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$ . Expected winnings $= \dfrac{1}{6} \times 10 + \dfrac{5}{6} \times 0 = \dfrac{10}{6} \approx 1.67$ dollars. Expected profit $= 1.67 - 2 = -0.33$ dollars. The game is not fair — on average, the player loses about $33$ cents per game.
$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ . Since $P(A \text{ and } B) \geq 0$ , we have $P(A \text{ or } B) \leq P(A) + P(B)$ . Equality holds when $P(A \text{ and } B) = 0$ , i.e. when $A$ and $B$ are mutually exclusive.

Cards $1$ – $5$ : odd numbers are $1, 3, 5$ (three), even numbers are $2, 4$ (two). For an even sum, both cards must be the same parity. $P(\text{both odd}) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20}$ . $P(\text{both even}) = \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{2}{20}$ . $P(\text{even sum}) = \dfrac{6}{20} + \dfrac{2}{20} = \dfrac{8}{20} = \dfrac{2}{5}$ .
$P(\text{both red}) = \dfrac{n}{n+3} \times \dfrac{n-1}{n+2} = \dfrac{1}{2}$ . So $2n(n-1) = (n+3)(n+2)$ , giving $2n^2 - 2n = n^2 + 5n + 6$ , i.e. $n^2 - 7n - 6 = 0$ . Using the quadratic formula: $n = \dfrac{7 + \sqrt{49 + 24}}{2} = \dfrac{7 + \sqrt{73}}{2} \approx \dfrac{7 + 8.544}{2} \approx 7.77$ . Since $n$ must be a positive integer, we check $n = 8$ : $\dfrac{8}{11} \times \dfrac{7}{10} = \dfrac{56}{110} = \dfrac{28}{55} \neq \dfrac{1}{2}$ . Check $n = 6$ : $\dfrac{6}{9} \times \dfrac{5}{8} = \dfrac{30}{72} = \dfrac{5}{12} \neq \dfrac{1}{2}$ . Since no integer solution exists, the equation $n^2 - 7n - 6 = 0$ has no positive integer root. Revisiting: if we allow $P(\text{both red}) = \dfrac{1}{2}$ to be approximate, $n = 8$ gives $\dfrac{28}{55} \approx 0.509$ , which is closest. However, for an exact solution: no integer value of $n$ works — this demonstrates that not every target probability is achievable with whole numbers of marbles.
Team A wins in 2 games: $P = 0.6 \times 0.6 = 0.36$ . Team A loses game 1, wins games 2 and 3: $P = 0.4 \times 0.6 \times 0.6 = 0.144$ . Team A wins game 1, loses game 2, wins game 3: $P = 0.6 \times 0.4 \times 0.6 = 0.144$ . Total: $P(\text{A wins series}) = 0.36 + 0.144 + 0.144 = 0.648$ .
$P(\text{none solve}) = (1 - 0.7)(1 - 0.8)(1 - 0.9) = 0.3 \times 0.2 \times 0.1 = 0.006$ . $P(\text{at least one solves}) = 1 - 0.006 = 0.994$ .

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