Question

A pair of dice is rolled. Find the probability of rollinga) a sum not more than 3,b) a sum not less than 8,c) a sum between 6 and 11 (exclusive).

Answer 1

Answer:

Explanation:

Let's go ahead and list all the possible outcomes when a pair of dice is rolled;

$\begin{gathered} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{gathered}$

We can see from the above that the total number of possible outcomes is 36

a) We're asked to determine the probability of rolling a sum not more than 3.

Let's list all the rolls that can produce a sum not more than 3;

$(1,1),(1,2),\text{and (2,1)}$

We can now find the probability as seen below;

$\begin{gathered} P(a\text{ sum not more than 3) =}\frac{Number\text{ of favourable outcomes}}{\text{Total number of possible outcomes}} \\ =\frac{3}{36} \\ =\frac{1}{12} \end{gathered}$

b) To determine the probability of rolling a sum not less than 8, let's list all the rolls that can produce a sum not less than 8;

$\begin{gathered} (2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3), \\ (6,4),(6,5),(6,6) \end{gathered}$

We can now find the probability as seen below;

$P(\text{not l}ess\text{ than 8)}=\frac{15}{36}=\frac{5}{12}$

c) To determine the probability of rolling a sum between 6 and 11(exclusive), let's list all the rolls that can produce a sum 7, 8, 9, and 10;

$\begin{gathered} (1,6),(2,5),(2,6),(3,4),(3,5),(3,6),(4,3),(4,4),(4,5),(4,6),(5,2),(5,3), \\ (5,4),(5,5),(6,1),(6,2),(6,3),(6,4) \end{gathered}$

We can now find the probability as seen below;

$P(a\text{ sum betwe}en\text{ 6 and 11(exclusive)})=\frac{18}{36}=\frac{3}{6}=\frac{1}{2}$