Question

Suppose a person rolls two balanced dice twice in succession. Determine the probability that the sum of the two numbers that appear on each of the three rolls will be 7.

Short answer

The probability that on each of the three rolls, the sum of the two numbers that appear will be 7 is \(\frac{1}{{216}}\).

Step-by-step solution

Given information

Suppose that a person rolls two balanced dice three times in succession.

It is required to determine the probability that the sum of the two numbers that appear on each of the three rolls will be 7.

Computing the required probability

According to the multiplication rule, the total probability of two or more independent events is given by:

\({\bf{P}}\left( {{\bf{total}}} \right){\bf{ = P}}\left( {{{\bf{A}}_{\bf{1}}}} \right) \times {\bf{P}}\left( {{{\bf{A}}_{\bf{2}}}} \right) \times ... \times {\bf{P}}\left( {{{\bf{A}}_{\bf{n}}}} \right)\)

If two balanced dice are rolled, the sample space will consist\({6^2} = 36\)of outcomes.

The combinations that will result in a sum of 7 are:

\(\left( {1,6} \right),\left( {6,1} \right),\left( {2,5} \right),\left( {5,2} \right),\left( {3,4} \right),\left( {4,3} \right)\)

The number of possible outcomes that will result in a sum of 7 is 6.

So, the probability that the sum is 7 for the first roll is:

\(\begin{array}{c}P\left( {{A_1}} \right) = \frac{6}{{36}}\\ = \frac{1}{6}\end{array}\)

Similarly, for the second and third rolls, the probabilities are:

\(P\left( {{A_2}} \right) = P\left( {{A_3}} \right) = \frac{1}{6}\)

Thus, the probability that on each of the three rolls, the sum of the two numbers that appear will be 7 is:

\(\begin{aligned}{}P\left( {total} \right) &= P\left( {{A_1}} \right) \times P\left( {{A_2}} \right) \times P\left( {{A_3}} \right)\\ &= \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}\\ &= \frac{1}{{216}}\end{aligned}\)

Therefore, the required probability is \(\frac{1}{{216}}\) .