Question

Find the probability of getting a sum of $9$ from the two throws of a dice.
(a) $\dfrac{1}{9}$
(b) $\dfrac{1}{4}$
(c) $\dfrac{1}{3}$
(d) None of these

Answer 1

Hint: The given problem revolves around the concepts of probability. So, we will use the definition of probability for each outcome or for the given outcome particularly. Using the formula for the probability i.e., if the probability of any event A is denoted by ‘P(A)’ and the formula to find the probability is \[P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( s \right)}}\], where n(s) denotes the favorable outcomes and n(A) denotes the total number of outcomes for an respective event.

Complete step-by-step solution:
Since, the two dice is fair,
As a result, for two dice there will be total of $36$ possible outcomes of getting number for the play (from one to six particularly), (when drawn simultaneously or randomly) are equally probable that is,
\[ p\left( A \right) = \left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right), \\
  \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right), \\
  \left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right), \\
  \left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right), \\
  \left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right), \\
  \left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right) \]
Where, the first number shows the outcome on the first dice and the other number shows the outcome on the second dice respectively.
‘$p\left( A \right)$’ denotes the total (entire) probability of the event on both the dice respectively.
Hence, we know that probability of any event to find the required number of possibilities can be found by the following formula i.e.
\[\text{Probability} = \dfrac{\text{Number of outcomes}}{\text{Total outcomes in the event}} = \dfrac{{n\left( s \right)}}{{n\left( A \right)}}\] … ($1$)
Where, $n\left( s \right) = $ number of possible outcomes in the respective event/s,
$n\left( A \right) = $ Total number of outcomes in the event
From the above mentioned possibilities of the event there are total $36$ possibilities of outcomes which is mathematically expressed as,
$n\left( A \right) = 36$ … (i)
It seems that, in this case, we need to find the possibilities of obtaining the sum of $9$.
Hence, total possible outcomes is as follows i.e.
$n\left( s \right) = \left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right)$
$\therefore n\left( s \right) = 4$ …………………… (ii)
From (i) and (ii),
Equation ($1$) becomes,
\[\text{Probability} = \dfrac{4}{{36}}\]
Or, considering its simplest form (i.e. the least multiple), that is dividing the equation by $4$, we get
 \[\text{Probability} = \dfrac{1}{9}\]
\[\therefore \]The option (a) is correct.

Note: One should know the basic formula of probability while solving a question. We should also know the concepts of independent events and that their probabilities are not affected by each other’s occurrences. We should take care of the calculations so as to be sure of our final answer. Also, we should know that the sum of probabilities of all the possible events in a situation is one.