Question

If three dice are thrown simultaneously, then the probability of getting a score of 7 is
A.\[\dfrac{5}{{216}}\]
B.\[\dfrac{1}{6}\]
C.\[\dfrac{5}{{72}}\]
D.None of these

Answer 1

Hint: We have to find the probability of getting a score of 7 when three dice are thrown at a time. For this, first we need to find the total outcomes of three dice thrown at a time then by using the definition of probability and on further simplification we get the required probability of choosing a card.

Complete answer:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the probability of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
\[\text{Probability of event to happen} P\left( E \right) = \dfrac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\]
Consider the given question:
The three dice are thrown simultaneously then we have to find the probability of getting a score of 7
If the Three dice are thrown simultaneously, total number of outcomes\[ = {6^3} = 216\]
The possible outcomes to getting a score 7 is
\[\left\{ {\left( {1,1,5} \right)\left( {1,5,1} \right)\left( {5,1,1} \right)\left( {2,2,3} \right)\left( {2,3,2} \right)\left( {3,2,2} \right)\left( {3,3,1} \right)} \right.\left( {3,1,3} \right)\left( {1,3,3} \right)\left( {4,2,1} \right)\left( {1,4,2} \right)\left( {2,1,4} \right)\]\[\left. {\left( {1,2,4} \right)\left( {2,4,1} \right)\left( {4,1,2} \right)} \right\} = 15\]
By the definition of probability
\[ \Rightarrow \,\,P\left( {getting\,a\,score\,of\,7} \right) = \dfrac{{Total\,possible\,outcomes\,to\,get\,7}}{{Total{\text{ }}number{\text{ }}of\,outcomes}}\]
\[ \Rightarrow \,\,P\left( {square\,number} \right) = \dfrac{{15}}{{216}}\]
Divide Both numerator and denominator of RHS by 3, then we get
 \[\therefore \,\,\,\,\,P\left( {square\,number} \right) = \dfrac{3}{{72}}\]
Hence, the required probability is \[\dfrac{3}{{72}}\]
Therefore, option C is the correct answer.

Note:
The probability is a number of possible values. Candidates must know the knowledge of dice, there are six faces in a single dice and if possible outcomes will be 6, then the total outcomes of three dice are thrown simultaneously are \[6 \times 6 \times 6 = {6^3}\]. When we imagine the dice and its thrown the solution will be easy to solve.