Question

Question: Use both the sample space (2.4) and the sample space (2.5) to answer the following questions about a toss of two dice.

(a) What is the probability that the sum is ≥ 4?

(b) What is the probability that the sum is even?

(c) What is the probability that the sum is divisible by 3?

(d) If the sum is odd, what is the probability that it is equal to 7?

(e) What is the probability that the product of the numbers on the two dice is 12?

Short answer

(a) The probability that the sum is greater than or equal to 4 is $\frac{11}{12}$.

(b) The probability that the sum is even is $\frac{1}{2}$.

(c) The probability that the sum is divisible by 3 is$\frac{1}{3}$ .

(d) The probability that the sum is 7 is$\frac{1}{6}$ .

(e) The probability that the product of the numbers on the two dice is 12 is $\frac{1}{9}$.

Step-by-step solution

Definition of Probability

The chances of occurrence of events are defined as a probability. The value of probability is always less than or equal to 1. Also, in the case of certain events, the probability is 1 and in the case of impossible events, the probability is 0.

(a) Determination of the probability that the sum is greater than or equal to 4

Create the Sample Space for the experiment when two dice are rolled. When two dice are rolled, then there are 36 points in the sample space. So, the sample space for the given problem is as follows,

$\left(\begin{array}{cccccc}\left(1,1\right)& \left(1,2\right)& \left(1,3\right)& \left(1,4\right)& \left(1,5\right)& \left(1,6\right)\\ \left(2,1\right)& \left(2,2\right)& \left(2,3\right)& \left(2,4\right)& \left(2,5\right)& \left(2,6\right)\\ \left(3,1\right)& \left(3,2\right)& \left(3,3\right)& \left(3,4\right)& \left(3,5\right)& \left(3,6\right)\\ \left(4,1\right)& \left(4,2\right)& \left(4,3\right)& \left(4,4\right)& \left(4,5\right)& \left(4,6\right)\\ \left(5,1\right)& \left(5,2\right)& \left(5,3\right)& \left(5,4\right)& \left(5,4\right)& \left(5,6\right)\\ \left(6,1\right)& \left(6,2\right)& \left(6,3\right)& \left(6,4\right)& \left(6,5\right)& \left(6,6\right)\end{array}\right)$

Write the sample space with associated probabilities with each point for the sum obtained when two dice are rolled.

$\begin{array}{cccccccccccc}\begin{array}{l}Sample\\ Space\end{array}& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ probability& \frac{1}{36}& \frac{2}{36}& \frac{3}{36}& \frac{4}{36}& \frac{5}{36}& \frac{6}{36}& \frac{5}{36}& \frac{4}{36}& \frac{3}{36}& \frac{2}{36}& \frac{1}{36}\end{array}$

It can be observed that there are 33 points in the sample space that gives the sum greater than or equal to 4, thus the probability is $\frac{33}{36}$or $\frac{11}{12}$.

Thus, the probability that the sum is greater than or equal to 4 is $\frac{11}{12}$.

(b) Determination of the probability that the sum is even

Find the probability of getting an even sum by adding all the associated probabilities where the sum is even.

$\begin{array}{c}p=\frac{1}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}+\frac{3}{36}+\frac{1}{36}\\ =\frac{18}{36}\\ =\frac{1}{2}\end{array}$

Thus, the probability that the sum is even is$\frac{1}{2}$ .

(c) Determination of the probability that the sum is divisible by 3

Find the probability of getting a sum which is divisible by 3 by adding all the associated probabilities where sum is divisible by 3.

$\begin{array}{c}p=\frac{2}{36}+\frac{5}{36}+\frac{4}{36}+\frac{1}{36}\\ =\frac{12}{36}\\ =\frac{1}{3}\end{array}$

Thus, the probability that the sum is divisible by 3 is$\frac{1}{3}$ .

(d) Determination of the probability that the sum is 7

Find the probability of getting a sum 7 from the obtained sample space.

$\begin{array}{c}p=\frac{6}{36}\\ =\frac{1}{6}\end{array}$

Thus, the probability of getting a sum 7 is $\frac{1}{6}$.

(e) Determination of the probability that product of the numbers on the two dice is 12

The product of two number is 12 when the pairs of numbers are $\left(2,6\right)$and $\left(3,4\right)$.Thus the points in the sample space which gives 12 as the product are $\left(2,6\right),\left(6,2\right),\left(3,4\right),\left(4,3\right)$and each has a probability of $\frac{1}{36}$.

Find the probability of getting a product 12 by adding the probability of each point.

$\begin{array}{c}p=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\\ =\frac{4}{36}\\ =\frac{1}{9}\end{array}$

Thus, the required probability is $\frac{1}{9}$.