A fair coin has 2 equally likely outcomes: heads or tails.
There is 1 favourable outcome for heads.
So the probability is \(\dfrac{1}{2}\).

The possible outcomes are HH, HT, TH and TT.
There are 4 equally likely outcomes.
Only 1 outcome gives two heads.
So the probability is \(\dfrac{1}{4}\).

The possible outcomes are HH, HT, TH and TT.
Exactly one head happens in HT and TH.
That is 2 outcomes out of 4.
So the probability is \(\dfrac{2}{4}=\dfrac{1}{2}\).

No heads means all three tosses are tails.
The only successful outcome is TTT.
There are \(2^3=8\) equally likely outcomes in total.
So the probability is \(\dfrac{1}{8}\).

There are \(2^3=8\) equally likely outcomes.
The only outcome with no heads is TTT, so \(P(\text{no heads})=\dfrac{1}{8}\).
Therefore \(P(\text{at least one head})=1-\dfrac{1}{8}=\dfrac{7}{8}\).

The probability of a head is \(\dfrac{1}{2}\).
The probability of an even number on a die is \(\dfrac{3}{6}=\dfrac{1}{2}\).
Multiply the probabilities:
\(\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4}\).

The possible outcomes are HH, HT, TH and TT.
The same result happens in HH and TT.
That is 2 outcomes out of 4.
So the probability is \(\dfrac{1}{2}\).

More heads than tails means 2 heads or 3 heads.
For 2 heads there are 3 outcomes: HHT, HTH, THH.
For 3 heads there is 1 outcome: HHH.
So there are 4 successful outcomes out of 8 in total.
The probability is \(\dfrac{4}{8}=\dfrac{1}{2}\).

A fair die has 6 equally likely outcomes.
Only 1 outcome is a 5.
So the probability is \(\dfrac{1}{6}\).

The even numbers are 2, 4 and 6.
That is 3 favourable outcomes out of 6.
So the probability is \(\dfrac{3}{6}=\dfrac{1}{2}\).

Numbers greater than 4 are 5 and 6.
That is 2 favourable outcomes out of 6.
So the probability is \(\dfrac{2}{6}=\dfrac{1}{3}\).

A total of 7 can happen in 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
There are 36 equally likely outcomes in total.
So the probability is \(\dfrac{6}{36}=\dfrac{1}{6}\).

A double means both dice show the same number.
The doubles are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
That is 6 outcomes out of 36.
So the probability is \(\dfrac{1}{6}\).

Totals less than 5 are 2, 3 and 4.
Ways to get 2: 1
Ways to get 3: 2
Ways to get 4: 3
So there are \(1+2+3=6\) successful outcomes out of 36.
The probability is \(\dfrac{6}{36}=\dfrac{1}{6}\).

A number less than 3 means 1 or 2, so the probability is \(\dfrac{2}{6}=\dfrac{1}{3}\).
The probability of a head is \(\dfrac{1}{2}\).
Multiply:
\(\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{1}{6}\).

Exactly one 6 can happen in two ways: first roll is 6 and second is not 6, or first is not 6 and second is 6.
Probability = \(\dfrac{1}{6}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{1}{6}=\dfrac{10}{36}=\dfrac{5}{18}\).

There are 36 equally likely outcomes in total.
The same number happens in (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
That is 6 outcomes.
So the probability is \(\dfrac{6}{36}=\dfrac{1}{6}\).

The even numbers are 2, 4, 6, 8 and 10.
That is 5 favourable cards out of 10.
So the probability is \(\dfrac{5}{10}=\dfrac{1}{2}\).

Numbers greater than 7 are 8, 9 and 10.
That is 3 favourable cards out of 10.
So the probability is \(\dfrac{3}{10}\).

The prime numbers from 1 to 10 are 2, 3, 5 and 7.
That is 4 favourable cards out of 10.
So the probability is \(\dfrac{4}{10}=\dfrac{2}{5}\).

The factors of 12 that appear from 1 to 10 are 1, 2, 3, 4 and 6.
That is 5 favourable cards out of 10.
So the probability is \(\dfrac{5}{10}=\dfrac{1}{2}\).

Odd numbers less than 6 are 1, 3 and 5.
That is 3 favourable cards out of 10.
So the probability is \(\dfrac{3}{10}\).

The multiples of 3 from 1 to 12 are 3, 6, 9 and 12.
That is 4 favourable cards out of 12.
So the probability is \(\dfrac{4}{12}=\dfrac{1}{3}\).

The square numbers from 1 to 9 are 1, 4 and 9.
That is 3 favourable cards out of 9.
So the probability is \(\dfrac{3}{9}=\dfrac{1}{3}\).

Numbers greater than 3 are 4 and 5.
That is 2 favourable cards out of 5.
So the probability is \(\dfrac{2}{5}\).

There are \(3+2+5=10\) marbles in total.
There are 2 blue marbles.
So the probability is \(\dfrac{2}{10}=\dfrac{1}{5}\).

There are \(4+6=10\) marbles in total.
Not red means blue, so there are 6 favourable marbles.
So the probability is \(\dfrac{6}{10}=\dfrac{3}{5}\).

There are \(5+3+2=10\) marbles in total.
Red or yellow means \(5+2=7\) favourable marbles.
So the probability is \(\dfrac{7}{10}\).

There are 7 marbles in total.
Probability first marble is red = \(\dfrac{3}{7}\).
Then 2 red marbles remain out of 6 marbles, so probability second marble is red = \(\dfrac{2}{6}\).
Multiply:
\(\dfrac{3}{7}\times\dfrac{2}{6}=\dfrac{6}{42}=\dfrac{1}{7}\).

There are \(2+3+1=6\) marbles in total.
There is 1 green marble.
So the probability is \(\dfrac{1}{6}\).