Question

Three dice are thrown simultaneously. What is the probability of getting a total score of at least $5$ of the numbers at their tops?

Answer 1

In the given question , there are three dice thrown simultaneously , therefore the total sample space will be obtained by taking the product of the total number of outcomes of each die . Since , the favorable outcomes are large in number , therefore we will calculate the probability of getting the sum less than $5$ and then subtracting from $1$ , since the total probability of any event is $1$ . Hence , it will provide the required probability . We will use the formula for the probability which is $= \dfrac{{Number{\text{ of favorable outcomes}}}}{{Total{\text{ number of outcomes}}}}$ to find the probability .

Complete step-by-step solution:
Total number of outcomes $= 6 \times 6 \times 6$
$= 216$
Sample space for getting a total of less than $5$ = \left\\{ {\left( {1,1,1} \right){\text{ }}\left( {{\text{1,1,2}}} \right){\text{ }}\left( {{\text{1,2,1}}} \right){\text{ }}\left( {{\text{2,1,1}}} \right)} \right\\} .
Now probability of getting a total of less than $5$ $= \dfrac{{Number{\text{ of favorable outcomes}}}}{{Total{\text{ number of outcomes}}}}$
On putting the values we get ,
Probability of getting a total of less than $5$ $= \dfrac{4}{{216}}$
On solving we get ,
Probability of getting a total of less than $5$ $= \dfrac{1}{{54}}$ .
The required probability will be $= 1 - \text{Probability of getting a total of less than}{\text{ 5}}$
On solving we get ,
$= 1 - \dfrac{1}{{54}}$
On simplifying we get ,
$= \dfrac{{54 - 1}}{{54}}$
On solving we get ,
$= \dfrac{{53}}{{54}}$
Therefore , the probability of getting a total score of at least $5$ of the numbers at their tops $= \dfrac{{53}}{{54}}$ .

Note: Probability is an estimate of the likeliness of an event to occur . Many events cannot be predicted with total certainty. When the favorable outcomes are large in number try to solve the question by subtracting the probability which is left over from the total probability which is $= 1$ . In the event of simultaneously obtaining the product of total number outcomes from each die or any other object.