Question

Probability of throwing \[16\] in one throw with three dice is
1. \[\dfrac{1}{{36}}\]
2. \[\dfrac{1}{{18}}\]
3. \[\dfrac{1}{{72}}\]
4. \[\dfrac{1}{9}\]

Answer 1

Hint: Here we have to find the probability of getting the sum of numbers as\[16\] when three dice are thrown. To find the probability we should know the total number of possible outcomes. . Find the total number of outcomes of the given experiment. Find the number of favorable outcomes and hence apply the formula for probability. Also keep in mind that the probability of any event is always between \[0\] and \[1\] only.

Complete step-by-step solution:
Sample Space: The sample space associated with a random experiment is the set of all possible outcomes. An event is a subset of the sample space.
Event: An event E is said to occur on a particular trial of the experiment if the outcome observed is an element of the set E.
We know that Probability (event) \[ = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\]
To find the probability of a single event to occur, first, we should know the total number of possible outcomes
Just as one die has six outcomes and two dice have \[{6^2} = 36\] outcomes , the total experiment of rolling three dice has \[{6^3} = 216\] outcomes.
When three dice are thrown, total number of sample space \[ = {6^3} = 216\]
Favourable cases \[ = 6\] i.e. \[(4,6,6),(6,6,4),(6,4,6),(5,5,6),(6,5,5),(5,6,5)\]
Hence, we get the required probability \[ = \dfrac{6}{{216}} = \dfrac{1}{{36}}\]
Therefore option (1) is the correct answer .

Note: To solve such kinds of questions one must know how to find the total number of outcomes of the given event. Keep in mind that the probability of any event can be between \[0\] and \[1\] only. Probability of any event can never be greater than \[1\]. Probability of any event can never be negative. Find the number of favorable cases and total cases correctly.