Question

Boucher rolls two dice sequentially. What is the probability of rolling 4 first then a number less than 3?

• A. $\frac{1}{18}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{5}{18}$

Answer 1

The correct option is A $\frac{1}{18}$

$\underset{–}{\text{Sample Space Diagram}}$

$1^{\text{st}}$ column and $1^{\text{st}}$ row signify the numbers on first die and second die, respectively.

	$\text{1}$	$\text{2}$	$\text{3}$	$\text{4}$	$\text{5}$	$\text{6}$
$\text{1}$	1-1	1-2	1-3	1-4	1-5	1-6
$\text{2}$	2-1	2-2	2-3	2-4	2-5	2-6
$\text{3}$	3-1	3-2	3-3	3-4	3-5	3-6
$\text{4}$	4-1	4-2	4-3	4-4	4-5	4-6
$\text{5}$	5-1	5-2	5-3	5-4	5-5	5-6
$\text{6}$	6-1	6-2	6-3	6-4	6-5	6-6

In the sample space diagram, there are $6$ rows and $6$ columns.

$\therefore$ Total number of possible outcomes for rolling two dice sequentially= $6\times 6=36$

Let $\text{E}$ be an event of rolling $4$ first then a number less then $3$.

$\therefore$ Number of favorable outcomes for the event $\text{E}$ is $2$ as in the sample space, only $4-1$ and $4-2$ satisfy of rolling $4$ in first die and then a number less than $3$ in second die.

$\text{P(E)}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$\Rightarrow \text{P(E)}=\frac{2}{36}=\frac{1}{18}$

$\underset{–}{\text{Alternative method (Product rule):}}$

Sample space for rolling a fair die= $\{1,2,3,4,5,6\}$
$\therefore$ Total number of possible outcomes= $6$
$\bullet$ For rolling $4$, number of favorable outcomes= $1$
$\bullet$ For rolling a number less than $3$, number of favorable outcomes= $2$
$(\because 1\text{and}2\text{are there in the sample space which are less than}3)$

We know,

$\text{P(rolling 4 then a number less than 3) = P(rolling 4)}\times \text{P(rolling a number less than 3)}$

Now,
$\text{P(rolling 4)}=\frac{\text{Number of}4\text{in sample space}}{\text{Total number of numbers in sample space}}$

$\Rightarrow \text{P(rolling 4)}=\frac{1}{6}$

Next,
$\text{P(rolling a number less than 3)}=\frac{\text{Number of numbers less than}3\text{in sample space}}{\text{Total number of numbers in sample space}}$

$\Rightarrow \text{P(rolling a number less than 3)}=\frac{2}{6}=\frac{1}{3}$

$\therefore \text{P(rolling 4 then a number less than 3) = P(rolling 4)}\times \text{P(rolling a number less than 3)}$
$\Rightarrow \text{P(rolling 4 then a number less than 3)}=\frac{1}{6}\times \frac{1}{3}=\frac{1\times 1}{6\times 3}=\frac{1}{18}$


$\therefore$ The probability that Boucher rolls $4$ then rolls a number less than $3$ is $\frac{1}{18}$.