Question

Two dice are thrown simultaneously. Match the following events with their probabilities.

• A. $\frac{5}{36}$
• B. $\frac{1}{9}$
• C. $\frac{1}{6}$
• D. 0

Answer 1

When two dice are thrown simultaneously, the number of possible outcomes is
$6^2=36.$
The possible outcomes are shown in the table below.

	1	2	3	4	5	6
1	(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
2	(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
3	(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
4	(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
5	(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(5,6)
6	(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

$\text{Probablity of an event}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

1. Getting a sum of 8:
$\text{Number of favourable outcomes}=\left\{\right(2,6),(3,5),(4,4),(5,3),(6,2\left)\right\}=5\therefore \text{Probability}=\frac{5}{36}$

2. Getting a product of 6:
$\text{Number of favourable outcomes}=\left\{\right(1,6),(2,3),(3,2),(6,1\left)\right\}=4\therefore \text{Probability}=\frac{4}{36}=\frac{1}{9}$

3. Getting a doublet:
$\text{Number of favourable outcomes}=\left\{\right(1,1),(2,2),(3,3),(4,4),(5,5),(6,6\left)\right\}=6\therefore \text{Probability}=\frac{6}{36}=\frac{1}{6}$

4. From the table, we can see that the least sum obtained is 2.
Thus, the probability of getting a sum of 1 is 0.