4. A fair die is rolled twice in a row. What is the probability that the sum of the two numbers obtained is a prime number?
4. Let the event $A=$ \{the sum of the numbers in two consecutive rolls is a prime number\}.

Since each roll of the dice has 6 outcomes, there will be $6 \cdot 6=36$ outcomes in total.

$1 \text{ POINT}$

All possible outcomes can be written as ordered pairs where the first element is the number obtained in the first roll, and the second element is the number obtained in the second roll of the dice.

Favorable are all those outcomes where the sum of the two elements is a prime number, and these are:

$(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4)$,

$(4,1),(4,3),(5,2),(5,6),(6,1)$ and $(6,5)$.

3 POINTS

There are a total of 15 favorable outcomes.

$1 \text{ POINT}$

Thus, $P(A)=\frac{15}{36}=\frac{5}{12}=0.41 \overline{6} \approx 41.67 \%$

$1 \text{ POINT}$

TOTAL 6 POINTS

Note 1: Listing favorable outcomes does not necessarily have to be in the form of ordered pairs (it can, for example, be in the form of $1+1, \ldots$). For one missing favorable outcome, give 2 POINTS, for two or three missing, give 1 POINT, and for more than three missing, give 0 POINTS. Further, follow the error in determining the probability of the event.

Note 2: The final solution is recognized regardless of the format (fraction, decimal, or percentage).