Probability

Each coin flip is independent, so previous results have no effect: the next flip is still $\dfrac{1}{2}$. (Expecting tails to be "due" is the gambler's fallacy; $\dfrac{1}{32}$ is the probability of five heads in a row from the start, a different question.)

Complementary events sum to 1: $P(\text{no rain}) = 1 - 0.3 = 0.7$. (1.3 added instead of subtracting; a probability cannot exceed 1.)

There are 13 hearts in 52 cards: $P(\text{heart}) = \dfrac{13}{52} = \dfrac{1}{4}$ (each of the 4 suits is one quarter). ($\dfrac{1}{52}$ is one specific card; 13 is the count.)

Use the complement: $P(\text{no heads}) = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$, so $P(\text{at least one head}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$. ($\dfrac{1}{8}$ forgot to subtract from 1; $\dfrac{3}{8}$ is exactly one head.)

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

$P(A)$ includes the 'only A' region plus the intersection: $10 + 5 = 15$ out of 30, so $P(A) = \dfrac{15}{30} = \dfrac{1}{2}$. ($\dfrac{1}{3}$ used only the 'only A' region; $\dfrac{13}{30}$ is $P(B)$.)

$P(A \cap B)$ is the overlap, 5 out of 30: $\dfrac{5}{30} = \dfrac{1}{6}$. ($\dfrac{23}{30}$ is $P(A \cup B)$ ('or', not 'and'); 5 is the count.)

$P(A \cup B)$ counts only A, both, and only B: $10 + 5 + 8 = 23$ out of 30, so $\dfrac{23}{30}$. Equivalently $P(A) + P(B) - P(A \cap B) = \dfrac{15}{30} + \dfrac{13}{30} - \dfrac{5}{30} = \dfrac{23}{30}$. (1 wrongly included the 7 in 'neither'.)

$P(\text{both}) = \dfrac{30}{100} = \dfrac{3}{10}$. ($\dfrac{3}{5}$ is $P(\text{maths})$; $\dfrac{11}{10}$ added the two without removing the overlap, giving an impossible value above 1.)

Geometric probability is the favourable area divided by the total area: $\dfrac{\pi(5)^2}{10^2} = \dfrac{25\pi}{100} = \dfrac{\pi}{4}$. ($\dfrac{1}{4}$ forgot the $\pi$ from the circle's area.)

Mutually exclusive means the events cannot both happen. 'Greater than 4' is $\{5, 6\}$ and 'less than 3' is $\{1, 2\}$: no overlap. ('Even and 2' overlap (2 is even); 'odd and prime' overlap (3 and 5).)

For mutually exclusive events $P(A \cup B) = P(A) + P(B) = 0.3 + 0.4 = 0.7$ (since $P(A \cap B) = 0$). (0.12 multiplied them, which gives $P(A \cap B)$ for independent events.)

These events overlap, so use inclusion-exclusion: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.4 - 0.2 = 0.8$. (1.0 added without subtracting the overlap, the canonical error.)

The expected winnings are $0.3 \times \$5 = \$1.50$. Subtract the $\$2$ cost to play: $\$1.50 - \$2 = -\$0.50$, an average loss of 50 cents per game. ($\$1.50$ is the expected winnings before paying the entry cost.)

Without replacement, the second draw's probability depends on the first (the total and the colour counts change), so the events are dependent. (Independent would require with-replacement draws.)

Multiply along the TT branches: $P(TT) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$. ($\dfrac{1}{3}$ treated HT and TH as a single outcome; for sequential events you multiply along the branches.)

Of $2^3 = 8$ outcomes, exactly two heads occurs for HHT, HTH and THH: 3 paths, so $\dfrac{3}{8}$. ($\dfrac{1}{8}$ counted only one path; the head can fall in different positions.)

With replacement the draws are independent: $P(\text{both red}) = \dfrac{4}{10} \times \dfrac{4}{10} = \dfrac{16}{100} = \dfrac{4}{25}$. ($\dfrac{12}{100}$ used $\dfrac{4}{10} \times \dfrac{3}{10}$, which is for without replacement.)

Without replacement the second draw has one fewer red and one fewer total: $\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$. ($\dfrac{4}{25}$ used with replacement; $\dfrac{12}{100}$ reduced the numerator but not the denominator.)

Two paths give exactly one blue: $P(BR) = \dfrac{5}{10} \times \dfrac{5}{10} = \dfrac{1}{4}$ and $P(RB) = \dfrac{1}{4}$, so $P(\text{exactly one blue}) = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$. ($\dfrac{1}{4}$ counted only one path.)

The Teens-row, Coffee-column cell is 25 out of 100: $\dfrac{25}{100} = \dfrac{1}{4}$. 'And' means a single cell, not a row or column total. ($\dfrac{5}{9}$ used the coffee column total as denominator, which is the conditional $P(\text{teen} \mid \text{coffee})$.)

The expected number is the probability multiplied by the number of trials: $0.35 \times 200 = 70$. ($130$ is the expected number of non-red spins, $0.65 \times 200$.)

A and B are independent if and only if $P(A \cap B) = P(A) \times P(B)$. Here $0.4 \times 0.5 = 0.2$, which equals the given $P(A \cap B) = 0.2$, so yes, they are independent. (Mutually exclusive events with non-zero probabilities are never independent.)

$P(\text{red}) = \dfrac{5}{5 + n} = \dfrac{1}{3}$. Cross-multiplying gives $3 \times 5 = 5 + n$, so $15 = 5 + n$ and $n = 10$. ($15$ is the total number of marbles, not the number of blue ones.)

$P(\text{both chocolate}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$. ($\dfrac{25}{64}$ used with replacement, $\dfrac{5}{8} \times \dfrac{5}{8}$; $\dfrac{9}{64}$ is both vanilla with replacement.)