Question

Compare the chances of throwing 4 with one die, 8 with two dice, and 12 with three dice.

Answer 1

Probability may be simply defined as the chance of happening of a certain event.
For an event, there might be many possible outcomes. Probability is the chance of any event to take place.
Here we will use the formula of probability which states that
${\text{Probability of an event}} = \dfrac{{{\text{No of favourable outcomes}}}}{{{\text{Total Outcomes}}}}$
In case of dice, throwing two or three dices simultaneously can be treated as throwing a single dice two or three times.
Here we will use the cases of dice, whose formula is ${6^n}$ where n is the number of times the dice has been thrown.

Complete step-by-step answer:
Firstly, we will discuss the case of throwing one die.
When one die is thrown, $n = 1$
$\Rightarrow Total{\text{ }}number{\text{ }}of{\text{ }}outcomes = {6^1} = 6$
Out of these, only one is 4.
So, $Favourable{\text{ }}Outcomes = 1$
${\text{The probability of getting 4}} = \dfrac{{{\text{No of favourable outcomes}}}}{{{\text{Total Outcomes}}}}$
$\Rightarrow P(4) = \dfrac{1}{6}$
The probability of getting 4 with one die is $\dfrac{1}{6}$
Now, we will discuss the case of throwing two dice.
When two dice are thrown, $n = 2$
$\Rightarrow Total{\text{ }}number{\text{ }}of{\text{ }}outcomes = {6^2} = 36$
Out of these, we get 8 in only these cases = \left\\{ {\left( {2,{\text{ }}6} \right),\left( {3,{\text{ }}5} \right),\left( {4,{\text{ }}4} \right),\left( {5,{\text{ }}3} \right),\left( {6,{\text{ }}2} \right)} \right\\}
So, $Favourable{\text{ }}Outcomes = 5$
${\text{The probability of getting 8 on 2 dice}} = \dfrac{{{\text{No of favourable outcomes}}}}{{{\text{Total Outcomes}}}}$
$\Rightarrow P(8{\text{ }}on{\text{ }}2{\text{ }}dice) = \dfrac{5}{{36}}$
The probability of getting a sum of 8 with 2 die is $\dfrac{5}{{36}}$
Now, we will discuss the case of throwing three dice.
When three dice are thrown, $n = 3$
$\Rightarrow Total{\text{ }}number{\text{ }}of{\text{ }}outcomes = {6^3} = 216$
Out of these, we get 12 in only these cases =

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

So, $Favourable{\text{ }}Outcomes = 25$
${\text{The probability of getting 12 on 3 dice}} = \dfrac{{{\text{No of favourable outcomes}}}}{{{\text{Total Outcomes}}}}$
$\Rightarrow P(12{\text{ }}on{\text{ 3 }}dice) = \dfrac{{25}}{{216}}$
The probability of getting a sum of 12 with 3 dies is $\dfrac{25}{{216}}$

Note: Some additional information about probability is as follows:
The sum of probabilities of all the events of an experiment is always equal to 1.
Probability can have minimum value 0 and maximum value 1.
If the sum of probabilities of an experiment is either less than 0 or greater than 1, then there is a mistake in the calculation.
Probability of an event which is which will definitely take place is 1. Such an event is called a True event.
Probability of an event which has not taken place is always taken as 0. Such an event is called False event