Question

Two dice are thrown simultaneously. The probability of getting a total of at most $5$ of the number appearing on their top is:
(A) $\dfrac{1}{4}$
(B) $\dfrac{7}{{36}}$
(C) $\dfrac{5}{{18}}$
(D) $\dfrac{9}{{35}}$

Answer 1

When two dies are rolled together, we get $\left( {1,1} \right)$,$\left( {1,2} \right)$,$\left( {1,3} \right)$,$\left( {1,4} \right)$,$\left( {1,5} \right)$,$\left( {1,6} \right)$,$\left( {2,1} \right)$,$\left( {2,2} \right)$,$\left( {2,3} \right)$,$\left( {2,4} \right)$,$\left( {2,5} \right)$,$\left( {2,6} \right)$,$\left( {3,1} \right)$,$\left( {3,2} \right)$,$\left( {3,3} \right)$,$\left( {3,4} \right)$,$\left( {3,5} \right)$,$\left( {3,6} \right)$,$\left( {4,1} \right)$,$\left( {4,2} \right)$,$\left( {4,3} \right)$,$\left( {4,4} \right)$,$\left( {4,5} \right)$,$\left( {4,6} \right)$,$\left( {5,1} \right)$,$\left( {5,2} \right)$,$\left( {5,3} \right)$,$\left( {5,4} \right)$,$\left( {5,5} \right)$,$\left( {5,6} \right)$,$\left( {6,1} \right)$,$\left( {6,2} \right)$,$\left( {6,3} \right)$,$\left( {6,4} \right)$,$\left( {6,5} \right)$ and $\left( {6,6} \right)$ . Now choose the outcome where we get the sum as $2,3,4{\text{ or 5}}$ . Use the formula for probability to find the required answer.

Complete step-by-step answer:
Here in this problem, we are given that two dices are rolled and we need to find the probability of getting a sum of two obtained numbers on top as $5$ or less.
Before starting with the solution we need to understand the concept of probability first. 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. The value is expressed from zero to one. The meaning of probability is the extent to which something is likely to happen. This is the basic probability theory, which is also used in the probability distribution, where you will learn the possibility of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
$\Rightarrow$ Probability $= \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}$
When two dices are rolled together, with each having possibilities of $1,2,3,4,5{\text{ and }}6$.
$\left( {1,1} \right)$,$\left( {1,2} \right)$,$\left( {1,3} \right)$,$\left( {1,4} \right)$,$\left( {1,5} \right)$,$\left( {1,6} \right)$,$\left( {2,1} \right)$,$\left( {2,2} \right)$,$\left( {2,3} \right)$,$\left( {2,4} \right)$,$\left( {2,5} \right)$,$\left( {2,6} \right)$,$\left( {3,1} \right)$,$\left( {3,2} \right)$,$\left( {3,3} \right)$,$\left( {3,4} \right)$,$\left( {3,5} \right)$,$\left( {3,6} \right)$,$\left( {4,1} \right)$,$\left( {4,2} \right)$,$\left( {4,3} \right)$,$\left( {4,4} \right)$,$\left( {4,5} \right)$,$\left( {4,6} \right)$,$\left( {5,1} \right)$,$\left( {5,2} \right)$,$\left( {5,3} \right)$,$\left( {5,4} \right)$,$\left( {5,5} \right)$,$\left( {5,6} \right)$,$\left( {6,1} \right)$,$\left( {6,2} \right)$,$\left( {6,3} \right)$,$\left( {6,4} \right)$,$\left( {6,5} \right)$ and $\left( {6,6} \right)$
So, when two dices are rolled together then the total number of outcomes possible can be calculated by multiplying the number of choices on each of the places, i.e.
$\Rightarrow$ Total possible outcomes $= 6 \times 6 = {6^2} = 36$
Now let’s try to figure out the arrangements where the sum is lesser than or equal to $5$ , i.e. $2,3,4{\text{ or }}5$
$\Rightarrow$ $2 + 3 = 3 + 2 = 5{\text{ and }}4 + 1 = 1 + 4 = 5$ ; which gives us the sum of $5$ in four possible ways
$\Rightarrow 3 + 1 = 1 + 3 = 4{\text{ and }}2 + 2 = 4$ ; which gives us the sum of $4$ in three possible ways
$\Rightarrow 2 + 1 = 1 + 2 = 3{\text{ }}$ ; which gives us the sum of $3$ in two possible ways
$\Rightarrow 1 + 1 = 2$ ; which gives us the sum of $2$ in one possible way
Therefore, the total number of favourable outcomes will be $4 + 3 + 2 + 1 = 10$
Thus, by using the formula for probability, we get:
$\Rightarrow$ Probability $= \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{10}}{{36}} = \dfrac{5}{{18}}$
Hence, the option (C) is the correct answer.

Note: In probability, questions like this can be solved with careful arrangement of favourable outcomes. Be careful while finding them. Notice the sum of at most $5$ means the sum should be $5$ or less than $5$ , but we did not include the number $1$ because it is not possible to obtain it as the sum of two numbers on top of dice. Be careful while counting the favourable outcomes of the required event.