Question

In a single throw of two dice, the probability of getting more than 7 is
A. ${\displaystyle \frac{7}{36}}$
B. ${\displaystyle \frac{7}{12}}$
C. ${\displaystyle \frac{5}{12}}$
D. ${\displaystyle \frac{5}{36}}$

Answer 1

Hint:Here the given question is based on the concept of probability. We have to find the probability of getting more than 7 when two dice are thrown at a single time. For this, first we need to find the total outcomes of two 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 step by step 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.
$\text{Probability of event to happen}P\left(E\right)={\displaystyle \frac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}}$

Consider the given question: The two dice are thrown at a single time then we have to find the probability of getting a more than 7. If the Three dice are thrown simultaneously
$\begin{array}{rl}& \{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ & \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\\ & \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\\ & \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\\ & \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\\ & \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\}\end{array}$

Total number of outcomes$=36$
A sum of 7 can be obtained in 6 ways =$\left\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\right\}$
A sum of 8 can be obtained in 5 ways $=\left\{(2,6),(3,5),(4,4),(5,3),(6,2)\right\}$
A sum of 9 can be obtained in 4 ways $=\left\{(3,6),(4,5),(5,4),(6,3)\right\}$
A sum of 10 can be obtained in 3 ways $=\left\{(4,6),(5,5),(6,4)\right\}$
A sum of 11 can be obtained in 2 ways $=\left\{(5,6),(6,5)\right\}$
A sum of 12 can be obtained in 1 way $=\left\{(6,6)\right\}$
Therefore, a sum greater than 7 occurs when the total is 8,9,10,11 or 12.
So, the number of possible outcomes to getting a more than 7 is: $5+4+3+2+1=15$ ways.

By the definition of probability
$\Rightarrow P\left(\text{getting a score more than 7}\right)={\displaystyle \frac{\text{Total possible outcomes to get more than 7}}{\text{Total number of outcomes}}}$
$\Rightarrow P\left(\text{getting a score more than 7}\right)={\displaystyle \frac{15}{36}}$
Divide Both numerator and denominator of RHS by 3, then we get
$\therefore P\left(\text{getting a score more than 7}\right)={\displaystyle \frac{5}{12}}$
Hence, the required probability is ${\displaystyle \frac{5}{12}}$.

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 two dice thrown simultaneously are $6\times 6=36$. When we imagine the dice and its thrown the solution will be easy to solve.