Question

In a single throw of two dice, find the probability of getting a total of $10$, getting a total of $9$ or $11$, getting a sum greater than $9$, getting a doublet of even numbers and not getting the same number on the two dice.

Answer 1

Step 1: Find the number of possible outcomes:

Sample space for throwing two dice will be,

$$\begin{gathered} n\left(S\right)={\text{(number of possible outcomes of throwing a dice)}}^{\text{number of dices thrown}} \\ n\left(S\right)={\left(6\right)}^2 \\ n\left(S\right)=36 \end{gathered}$$

Hence, the total number of outcomes will be $36$

Step 2: Use the formula of Probability

$$\begin{gathered} \text{Probability =}\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \\ \text{P(A)=}\frac{n\left(A\right)}{n\left(S\right)} \end{gathered}$$

Step 3: Finding the probability of getting a total of $10$

The favorable outcomes of getting a total of $10$ $=\left\{(4,6),(5,5),(6,4)\right\}$

$\Rightarrow n\left(A\right)=3$

Hence, probability will be,

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

Step 4: Finding the probability of getting a total of $9or11$:

The favorable outcomes of getting a total of $11$$=\left\{(5,6),(6,5)\right\}$

The favorable outcomes of getting a total of $9$ $=\left\{(4,5),(5,4),(6,3),(3,6)\right\}$

$n\left(A\right)=2+4=6$

Hence, probability will be,

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

Step 5: Finding the probability of getting a sum greater than 9:

The favorable outcomes of getting a sum greater than $9$ will be,

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

$n\left(A\right)=6$.

Thus, the probability will be,

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

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

The favorable outcomes of getting a doublet of even numbers will be,

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

$n\left(A\right)=3$

Thus, the probability will be,

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

Step 7: Finding the probability of not getting the same number on the two dice:

The favorable outcomes of getting the same number on the two dice will be,

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

$n\left(A\right)=36-6=30$

Thus, the probability will be,

$$\begin{gathered} \text{P(A)=}\frac{30}{36} \\ =\frac{5}{6} \end{gathered}$$

Therefore, the probability of getting a total of $10=\frac{1}{12}$, getting a total of $9$ or $11=\frac{1}{6}$, getting a sum greater than $9=\frac{1}{6}$, getting a doublet of even numbers $=\frac{1}{12}$ and not getting the same number on the two dice $=\frac{5}{6}$