Question

find the probability the roll was a
(a) sum of 12.
the probability the roll was a sum of 12 is\\( \frac { 1 } { 36 } \\).
(b) sum that is even.
the probability the roll was a sum that is even is\\( \frac { 1 } { 2 } \\).
(c) four on one or both dice.
the probability the roll was a four on one or both dice is\\( \square \\)

Explanation:

Step1: Count total outcomes

When two dice are rolled, the total number of outcomes is \(n(S)=6\times6 = 36\) (using the fundamental counting principle).

Step2: Count favorable outcomes for sum of 4

The pairs \((1,3)\), \((2,2)\), \((3,1)\) give a sum of 4. So \(n(A)=3\).

Step3: Calculate probability

The probability formula is \(P(A)=\frac{n(A)}{n(S)}\). Substituting \(n(A) = 3\) and \(n(S)=36\), we get \(P(A)=\frac{3}{36}=\frac{1}{12}\).

Answer:

\(\frac{1}{12}\)