Question

In a single throw of a pair of dice, the probability of getting the sum a perfect square is
A) \dfrac{1}{{18}} \\\
B) \dfrac{7}{{36}} \\\
C) \dfrac{1}{6} \\\
D) $\dfrac{2}{9} \\\$

Answer 1

In this question, we have to find out the probability of getting the sum of a perfect square. We know the sum varies from $2$ to $12$ if two dices are thrown because the minimum will be $2$ if both dices give outcomes $1,1$ and $12$ if outcomes are $6,6$. Note the perfect squares come between $2$ to $12$ and work out the cases about the outcomes which will give the sums as perfect squares. After doing so, find out the probability by using (Favorable outcomes/Total outcomes).

Complete step-by-step answer:
If two dies are thrown, then their sum will vary from $2$ to $12$, i.e. sum $E\left[ {2,12} \right]$
Now, between $2$ to $12$, the perfect squares are $\to 4,9$
Now, the sum will be $4$, if the outcomes are –
$4E\left( {3,1} \right),\left( {2,2} \right),\left( {1,3} \right)$
& sum will be $9$, if outcomes will –
$9E\left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right)$.
Now, we know the total outcomes of dice, when they are thrown.
Outcomes are $6$ when single dice is thrown
Then, total outcomes are $6 \times 6 = 36$.
Total outcomes $= 36$.
$\therefore$Probability of getting sum of a perfect square –
P (sum=perfect square) = favorable outcomes/total outcomes
$= \dfrac{{\left( {sum = 4} \right)and\left( {sum = 9} \right)}}{{36}}$
$= \dfrac{{3 + 4}}{{36}}$
$= \dfrac{7}{{36}}$ [ outcomes for $4$ as sum $= 3$ & for $9$ is $4$].
$\therefore$Required probability is $\dfrac{7}{{36}}$ $\left( {option \to b} \right)$

Note: To solve the questions regarding probability, it is better to note down the outcomes of the questions asked in the solution, a set of outcomes are found which gives the sum of $4$ & $9$. It makes the question easier to find probability as it is known that total outcomes are $36$ when two dice are thrown and favorable conditions are solved. Their division gives out the required probability.