Three different solutions are given below.  Let X be the random variable representing the # of fives.

Method 1:  Analysing the sample space

“At least one five” means one or more fives – or specifically for this question – one or two fives. From the table (left) depicting the sample space, it can be seen that there are ten outcomes with one five and one outcome with two fives (shown in red). Therefore, the probability of getting at least one five when rolling a die twice is \({\textrm{P}}\left( {X \ge 1} \right) = \frac{{11}}{{36}}\).
 

Method 2:  Adding probabilities of all the outcomes that make up the event

The ‘event’ of obtaining at least one five consists of three different outcomes: (1) a 5 on the 1st die, and not a 5 on the 2nd die, (2) not a 5 on the 1st die, and a 5 on the 2nd die, and (3) a 5 on both dice [note: 5´ means “not 5”]

\({\textrm{P}}\left( {5,5'} \right) = \frac{1}{6} \cdot \frac{5}{6} = \frac{5}{{36}}\)            \({\textrm{P}}\left( {5',5} \right) = \frac{5}{6} \cdot \frac{1}{6} = \frac{5}{{36}}\)           \({\textrm{P}}\left( {5,5} \right) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{{36}}\)

From these three probabilities, it follows that \({\textrm{P}}\left( {X \ge 1} \right) = \frac{5}{{36}} + \frac{5}{{36}} + \frac{1}{{36}} = \frac{{11}}{{36}}\)

Method 3:  Complement approach

\({\textrm{P}}\left( {X \ge 1} \right) + {\textrm{P}}\left( {X = 0} \right) = 1\;\;\;\; \Rightarrow \;\;\;\;{\textrm{P}}\left( {X \ge 1} \right) = 1 - {\textrm{P}}\left( {X = 0} \right)\)

\({\textrm{P}}\left( {X = 0} \right)\) is the probability of not getting a 5 on either of the dice.

Therefore, \({\textrm{P}}\left( {X \ge 1} \right) = 1 - \frac{5}{6} \cdot \frac{5}{6} = 1 - \frac{{25}}{{36}} = \frac{{11}}{{36}}\)

Clearly, the complement approach is the most efficient way of computing this particular probability. This is often the case with any question involving the phrase "at least one" – it's worth being explicit about this with students: "whenever you see 'at least one, try using the complement'."

Students should be presented with several examples that illustrate the usefulness of a complement approach in a variety of probability questions. See the page IA idea #3 - Choosing with replacement in my Exploration (IA) Ideas section.