Question

Two fair dice are rolled simultaneously. The probability of getting $2$, $2$ is .

• A. $\frac{1}{30}$
• B. $\frac{1}{6}$
• C. $\frac{1}{36}$

Answer 1

The correct option is C $\frac{1}{36}$

$\underset{–}{\text{Solving by sample space diagram}}$

Sample space diagram for rolling two dice:
$$\left\{\begin{array}{cccccc}(1,1)& (1,2)& (1,3)& (1,4)& (1,5)& (1,6)\\ (2,1)& (2,2)& (2,3)& (2,4)& (2,5)& (2,6)\\ (3,1)& (3,2)& (3,3)& (3,4)& (3,5)& (3,6)\\ (4,1)& (4,2)& (4,3)& (4,4)& (4,5)& (4,6)\\ (5,1)& (5,2)& (5,3)& (5,4)& (5,5)& (5,6)\\ (6,1)& (6,2)& (6,3)& (6,4)& (6,5)& (6,6)\end{array}\right\}$$In the above table, row $1$ depicts the numbers on one die and column $1$ depicts the numbers on other die.

For possible outcomes, we have $6$ rows and $6$ columns in the sample space.

Total number of possible outcomes= $6\times 6=36$
Number of favorable outcomes for rolling $2,2$= $1$ (see $2^{\text{nd}}$ row and $2^{\text{nd}}$ column in the sample space diagram)

$\therefore \text{P(rolling 2 and 2)}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$\Rightarrow \text{P(rolling 2 and 2)}=\frac{1}{36}$

$\underset{–}{\text{Solving by multiplication of single events}}$

It is clear that simultaneous rolling of $2,2$ in two fair dice is an independent event, i.e., the rolling of $2$ on one die does not have any impact on rolling of $2$ on the other die.

$\therefore \text{P(rolling 2 and 2)=P(rolling 2)}\times \text{P(rolling 2)}$

$\Rightarrow \text{P(rolling 2 and 2)}=\frac{1}{6}\times \frac{1}{6}$

($\because \text{P(rolling 2)}=\frac{1}{6}$ for each die)

$\Rightarrow \text{P(rolling 2 and 2)}=\frac{1\times 1}{6\times 6}=\frac{1}{36}$

$\therefore$ Using both the above methods, we get the probability of rolling $2,2$ when two fair dice are rolled simultaneously is $\frac{1}{36}$.