Question

In a single throw of two dice, find the probability of: getting a total of $10$, getting a sum greater than$9$, getting a total of $9$or $11$, getting a doublet, getting a doublet of even numbers

Answer 1

Step 1: Use the formula of probability

$P\left(A\right)=\frac{\text{number of outcome favorable to A}}{\text{total number of Possible outcome}}$

or

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

where,

$P\left(A\right)$ is probability of an event ‘$A$’

$n\left(A\right)$ is the favorable outcome of the event ‘$A$’

$n\left(S\right)$ is total number of possible outcomes in a sample space

Step 2:Find the probability of getting a total of $10$

$S=\left\{\right(1,1),(1,2),...,(6,6\left)\right\}$

$S$ is a sample space of all the possible outcomes which is $36$

Thus, total number of possible outcomes in a sample space is

$n\left(S\right)=36$

In these $36$ outcomes three contain a total of $10$

$A=\left\{\right(4,6),(5,5),(6,4\left)\right\}$

Thus, the favorable outcome of the event ‘$A$’ is $n\left(A\right)=3$

$$\begin{gathered} P\left(A\right)=\frac{3}{36} \\ =\frac{1}{12} \end{gathered}$$

Step 3: Find the probability of getting a greater than $9$

Let $A$ be the event of "getting a sum greater than $9$"

In the $36$ outcomes six contain a sum greater than $9$

$A=\left\{\right(4,6),(5,5),(6,4),(5,6),(6,5),(6,6\left)\right\}$

Thus, the favorable outcome of the event ‘$A$’ is $n\left(A\right)=6$

$$\begin{gathered} P\left(A\right)=\frac{6}{36} \\ =\frac{1}{6} \end{gathered}$$

Step 4:Find the probability of getting a total $9$ or $11$

Let $A$ be the event of "getting a total $9$ or $11$"

In the $36$ outcomes six contain a total $9$ or $11$

$A=\left\{\right(4,5),(5,4),(5,6),(3,6),(6,3),(6,5\left)\right\}$

Thus, the favorable outcome of the event ‘$A$’ is $n\left(A\right)=6$

$$\begin{gathered} P\left(A\right)=\frac{6}{36} \\ =\frac{1}{6} \end{gathered}$$

Step 5: Find the probability of getting a doublet

Let $A$ be the event of "getting a doublet"

In the $36$ outcomes six contain a doublet

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

Thus, the favorable outcome of the event ‘$A$’ is $n\left(A\right)=6$

$$\begin{gathered} P\left(A\right)=\frac{6}{36} \\ =\frac{1}{6} \end{gathered}$$

Step 6: Find the probability of getting a doublet of even numbers

Let $A$ be the event of "getting a doublet of even numbers"

In the $36$ outcomes three contain a doublet of even numbers

$A=\left\{\right(2,2),(4,4),(6,6\left)\right\}$

Thus, the favorable outcome of the event ‘$A$’ is $n\left(A\right)=3$

$$\begin{gathered} P\left(A\right)=\frac{3}{36} \\ =\frac{1}{12} \end{gathered}$$

Hence, the probability of getting a total of 10 is $\frac{1}{12}$, getting a sum greater than 9 is $\frac{1}{6}$, getting a total of 9 or 11 is $\frac{1}{6}$, getting a doublet is $\frac{1}{6}$, getting a doublet of even numbers are $\frac{1}{12}$.