Conditional Probability and Independence

$P(\text{both green}) = \dfrac{5}{12} \times \dfrac{4}{11} = \dfrac{20}{132} = \dfrac{5}{33}$ (the second draw has one fewer green and one fewer total). ($\dfrac{25}{144}$ treats it as with replacement; always simplify the final fraction.)

Use the complement: $P(\text{at least one}) = 1 - P(\text{no reds}) = 1 - (0.7)^3 = 1 - 0.343 = 0.657$. ($0.343$ is $P(\text{no reds})$; $0.9$ wrongly adds the three probabilities.)

Each toss is independent, so previous results have no effect: the next toss is still $\dfrac{1}{2}$. (Expecting tails to be "due" is the gambler's fallacy; $\dfrac{1}{64}$ is the probability of six heads in a row beforehand, a different question.)

The two experiments are independent: $P(>4) = \dfrac{2}{6} = \dfrac{1}{3}$ (the $5$ or $6$) and $P(\text{tail}) = \dfrac{1}{2}$, so $\dfrac{1}{3} \times \dfrac{1}{2} = \dfrac{1}{6}$. ($\dfrac{1}{3}$ forgot the coin.)

Mutually exclusive, so add: $\dfrac{12}{30} + \dfrac{8}{30} = \dfrac{20}{30} = \dfrac{2}{3}$. ($\dfrac{20}{30}$ is correct but not simplified; always simplify.)

With replacement the draws are independent: $\dfrac{4}{10} \times \dfrac{4}{10} = \dfrac{16}{100} = \dfrac{4}{25}$. ($\dfrac{4}{10}$ is just one draw.)

Without replacement, adjust stage two: $\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$. ($\dfrac{4}{25}$ treats it as with replacement.)

Turn the probability into an equation: $P(\text{red}) = \dfrac{5}{5 + n} = \dfrac{1}{3}$, so $5 + n = 15$ and $n = 10$.

Three orderings $(HHT, HTH, THH)$ out of $2^3 = 8$: $P = \dfrac{3}{8}$. (One ordering alone gives $\dfrac{1}{8}$.)

Geometric probability is the favourable area over the total area: $\dfrac{\pi(5)^2}{20^2} = \dfrac{25\pi}{400} = \dfrac{\pi}{16}$.

The intersection cell (likes Maths and likes Science) is $20$. ($30$ and $35$ are marginal totals; $10$ is "Maths but not Science".)

The expected number of successes is the probability times the number of trials: $0.35 \times 200 = 70$.

The expected winnings are $0.3 \times \$5 = \$1.50$. Subtracting the $\$2$ cost gives an expected profit of $\$1.50 - \$2 = -\$0.50$ (an average loss of $50$ cents).

Restrict to the "likes Maths" row ($30$ students): $P(S \mid M) = \dfrac{20}{30} = \dfrac{2}{3}$. ($\dfrac{20}{35}$ conditions on Science instead, giving $P(M \mid S)$.)

The ratio $3 : 2$ over $15$ balls gives $9$ red and $6$ blue. Without replacement: $\dfrac{9}{15} \times \dfrac{8}{14} = \dfrac{72}{210} = \dfrac{12}{35}$. (First use the ratio to find the counts, then apply probability.)

Since $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ and $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$, they share the same numerator but differ when $P(A) \ne P(B)$. So in general the two conditionals are not equal (confusing them is the "base-rate" error).

$P(A \mid B) = \dfrac{0.2}{0.5} = 0.4$. ($2.5$ inverted the formula; a conditional probability is between $0$ and $1$.)

Independent events multiply, so the probability is $0.8^4 = 0.8 \times 0.8 \times 0.8 \times 0.8 = 0.4096$.

Addition rule with overlap: $P(M \cup D) = \dfrac{15}{30} + \dfrac{18}{30} - \dfrac{8}{30} = \dfrac{25}{30} = \dfrac{5}{6}$. Or count the Venn regions $7 + 8 + 10 = 25$. ($\dfrac{33}{30}$ forgot to subtract the overlap and exceeds $1$, which is impossible.)

$P(B \mid C) = \dfrac{0.25}{0.60} = \dfrac{25}{60} = \dfrac{5}{12} \approx 0.417$. ($\dfrac{5}{8} = 0.625$ inverted to $P(C \mid B)$.)

For independent events $P(A \cap B) = 0.6 \times 0.5 = 0.3$. Then $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.5 - 0.3 = 0.8$. ($1.1$ forgot to subtract the overlap, which is impossible for a probability.)

Test the identity: $P(A) \times P(B) = 0.4 \times 0.5 = 0.2 = P(A \cap B)$, so they are independent. (Independent events can have different probabilities and can overlap, so the "No" reasons are false.)

The favourable sector spans $210^\circ - 90^\circ = 120^\circ$ out of the full $360^\circ$: $\dfrac{120}{360} = \dfrac{1}{3}$. (Geometric probability using angle as the measure.)

Two paths give exactly one red: $P(RB) = \dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56}$ and $P(BR) = \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}$. Add them: $\dfrac{15}{56} + \dfrac{15}{56} = \dfrac{30}{56} = \dfrac{15}{28}$. ($\dfrac{15}{56}$ is only one path; $\dfrac{15}{64}$ assumes with replacement.)

Use $100$ students: $40$ study French, of whom $0.8 \times 40 = 32$ study Spanish; $60$ do not, of whom $0.3 \times 60 = 18$ study Spanish. Spanish total $32 + 18 = 50$. So $P(F \mid S) = \dfrac{32}{50} = \dfrac{16}{25} = 0.64$. ($0.80$ is the reverse $P(S \mid F)$.)