Question

 A die is thrown 4 times. The probability of getting perfect square in at least one throw is:
A. $\dfrac{16}{81}$
B. $\dfrac{65}{81}$
C. $\dfrac{23}{81}$
D. $\dfrac{55}{81}$

Answer 1

Hint: We are going to find the probability of getting a perfect square in at least one throw of a die by subtracting the probability of not getting a perfect square not in any throw from 1.

Complete step by step answer:
We know that the value of probability is from 0 to 1.
Number of possible outcomes when a die is thrown once is 6 (1, 2, 3, 4, 5, and 6).
Now, to get a perfect square as an outcome, two outcomes are possible (1, 4) in one throw of a die.
To not get a perfect square as an outcome, 4 outcomes are possible (2, 3, 5, 6) in one throw of a die.
Probability of any outcome on a die =$\dfrac{\text{Favourable outcome}}{\text{Total outcomes}}$
Now, we have to find the probability of getting a perfect square in at least one throw when a die is thrown four times.
Probability of getting a perfect square in at least one throw = 1 – (Probability of not getting a perfect square in any throw)
Probability of not getting a perfect square in one throw is$\dfrac{4}{6}$.
Probability of not getting a perfect square in 4 throws of a die is$\dfrac{4}{6}\times \dfrac{4}{6}\times \dfrac{4}{6}\times \dfrac{4}{6}$.
Using the above relation to find the probability of getting a perfect square in at least one throw is:
=$\begin{align}
  & 1-\left( \dfrac{4}{6}\times \dfrac{4}{6}\times \dfrac{4}{6}\times \dfrac{4}{6} \right) \\
 & =1-\left( \dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3} \right) \\
 & =1-\left( \dfrac{16}{81} \right) \\
 & =\dfrac{65}{81} \\
\end{align}$
Hence, the correct option is (b).

Note: The alternative way of solving the above problem is to find the probability of getting a perfect square once, twice, thrice and four times and then add all the cases but it will be a lengthy process so it’s better to solve the problem as discussed in the solution part.