Question

Two dice are thrown. Use the sample space (2.4) to answer the following questions.

(a) What is the probability of being able to form a two-digit number greater than

33 with the two numbers on the dice? (Note that the sample point 1, 4 yields

the two-digit number 41 which is greater than 33, etc.)

(b) Repeat part (a) for the probability of being able to form a two-digit number

greater than or equal to 42.

(c) Can you find a two-digit number (or numbers) such that the probability of

being able to form a larger number is the same as the probability of being able

to form a smaller number? [See note part (a)]

Short answer

(a) The probability of being able to form a two-digit number greater than 33 with the two numbers on the dice is $\frac{7}{12}$.

(b) The probability of being able to form a two-digit number greater than or equal to 42 with the two numbers on the dice is$\frac{17}{36}$ .

(c) A two-digit number such that the probability of being able to form a larger number is the same as the probability of being able to form a smaller number are all numbers in the sample space except $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)$.

Step-by-step solution

Definition of Uniform and Non-Uniform sample space

The uniform sample space of an experiment is the set of outcomes having the similar probability of occurring.

The non-uniform sample space of an experiment is the set of all possible mutually exclusive events that is each point has a different probability.

(a) Determination of the probability of being able to form a two-digit number greater than 33 with the two numbers on the dice 

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,

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

It can be observed from the sample space that there are 21 points in the sample space that gives a number greater than 33, thus the probability that a two-digit number is greater than 33 with the two numbers on the dice is formed asor $\frac{21}{36}$or $\frac{7}{12}$.

Thus, the required probability is $\frac{7}{12}$.

(b) Determination of the probability of being able to form a two-digit number greater than or equal to 42 with the two numbers on the dice

It can be observed that from the sample space there are 17 points in the sample space that gives a number greater than or equal to 42, thus the probability that a two-digit number is greater than or equal to 42 with the two numbers on the dice is formed is $\frac{17}{36}$.

Thus, the required probability is$\frac{17}{36}$ .

(c) Determination of a two-digit number (or numbers) such that the probability of being able to form a larger number is the same as the probability of being able to form a smaller number

It can be observed from the sample space that all numbers except $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)$can form a larger number as well as a smaller number with the same probability.

Thus, the required numbers are other than $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)$.