Probability

There is one "3" out of six equally likely outcomes: $P(3) = \dfrac{1}{6}$. (3 is an outcome, not a probability; every probability is between 0 and 1.)

$P(\text{green}) = \dfrac{6}{4 + 6} = \dfrac{6}{10} = \dfrac{3}{5}$. ($\dfrac{2}{5}$ is P(red); $\dfrac{6}{4}$ is the green-to-red ratio, not a probability.)

A certain event has probability 1; an impossible event has probability 0; an even chance is 0.5. (Probabilities are never above 1, so 100 is impossible.)

An event and its complement have probabilities that add to 1: $P(\text{no rain}) = 1 - 0.3 = 0.7$. (Adding to get 1.3 is impossible for a probability.)

$\dfrac{1}{4} = 0.25 = 25\%$. (75% is the complement; probabilities are often written as fractions, decimals or percentages.)

Greater than 4 means 5 or 6: 2 outcomes out of 6, so $\dfrac{2}{6} = \dfrac{1}{3}$. (Including 4 wrongly gives $\dfrac{1}{2}$; "greater than" excludes 4.)

BANANA has 6 letters, and A appears 3 times: $P(A) = \dfrac{3}{6} = \dfrac{1}{2}$. (3 is the count of A's, not the probability.)

Each coin has 2 outcomes, so together there are $2\times 2 = 4$: HH, HT, TH, TT. (The coins are distinguishable, so HT and TH are different outcomes, giving 4 not 3.)

Out of the four outcomes HH, HT, TH, TT, only HH is two heads: $\dfrac{1}{4}$. (Or $\dfrac{1}{2}\times\dfrac{1}{2} = \dfrac{1}{4}$.)

Not white $= 5 + 3 = 8$ out of $5 + 3 + 2 = 10$, so $\dfrac{8}{10} = \dfrac{4}{5}$. (Or use the complement: $1 - \dfrac{2}{10} = \dfrac{8}{10}$.)

There are 13 hearts in 52 cards: $\dfrac{13}{52} = \dfrac{1}{4}$ (there are four equal suits). ($\dfrac{1}{13}$ is the chance of a particular value such as an ace.)

Convert all to decimals: $\dfrac{1}{3}\approx 0.33$, $\dfrac{1}{4} = 0.25$, $0.6$, $30\% = 0.3$, $0.4$. The largest is $0.6$. (Always convert to a common form before comparing.)

All the probabilities must add to 1: $P(\text{green}) = 1 - 0.5 - 0.3 = 0.2$.

Expected number $= P \times \text{total} = 0.25 \times 200 = 50$. (This is a long-run expectation, not a guarantee of exactly 50.)

Each die has 6 outcomes, so together there are $6\times 6 = 36$ equally likely outcomes. (12 comes from adding $6+6$; 11 is the number of possible totals, not outcomes.)

Totals of 7 are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$: 6 outcomes out of 36, so $\dfrac{6}{36} = \dfrac{1}{6}$. (7 is the most likely total when rolling two dice.)

$\dfrac{3}{10}$ of 40 is $\dfrac{3}{10}\times 40 = 12$ blue marbles. (28 is the number that are not blue.)

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; $\dfrac{1}{32}$ is the chance of getting five heads in a row from the start.)

Students who play at least one sport $= 18 + 12 - 5 = 25$ (subtract the 5 counted twice). So neither $= 30 - 25 = 5$. (Adding 18 and 12 without subtracting the overlap double-counts the 5.)

Even numbers: 2, 4, 6, 8. Greater than 5: 6, 7, 8. Listing them once each gives 2, 4, 6, 7, 8, which is 5 outcomes: $\dfrac{5}{8}$. (Adding 4 and 3 double-counts 6 and 8.)

Mutually exclusive events share no outcomes. Even numbers are 2, 4 and 6, and "a 3" is odd, so they cannot both occur, making them mutually exclusive. (In every other pair, the events share an outcome, e.g. 'even' and 'a 2' both include 2.)

Experimental probability $= \dfrac{\text{heads}}{\text{trials}} = \dfrac{30}{50} = \dfrac{3}{5}$. (The theoretical value for a fair coin is $\dfrac{1}{2}$, but the experiment gave $\dfrac{3}{5}$.)

If $P(\text{red}) = \dfrac{1}{4}$, then for every 1 red there are 3 blue (a 1 : 3 ratio). With 9 blue, the number of red is $\dfrac{9}{3} = 3$. (Check: $\dfrac{3}{3 + 9} = \dfrac{3}{12} = \dfrac{1}{4}$.)

The two events are independent, so multiply: $P(\text{head})\times P(6) = \dfrac{1}{2}\times\dfrac{1}{6} = \dfrac{1}{12}$. (The sample space has $2\times 6 = 12$ equally likely outcomes, only one of which is head-and-6.)

The outcomes are HH, HT, TH, TT. "At least one head" is every outcome except TT: 3 out of 4, so $\dfrac{3}{4}$. (Or use the complement: $1 - P(TT) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.)