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(B) The last digit of the square of an integer depends only on the last digit of the integer itself.Let the last digit of the integer be $x$,where $x

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$\frac{3}{10}$

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(B) The last digit of the square of an integer depends only on the last digit of the integer itself.Let the last digit of the integer be $x$,where $x \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.The squares of these digits are:$0^2 = 0$$1^2 = 1$$2^2 = 4$$3^2 = 9$$4^2 = 16$ (last digit $6$)$5^2 = 25$ (last digit $5$)$6^2 = 36$ (last digit $6$)$7^2 = 49$ (last digit $9$)$8^2 = 64$ (last digit $4$)$9^2 = 81$ (last digit $1$)The last digits of the squares are $\{0, 1, 4, 9, 6, 5, 6, 9, 4, 1\}$.We want the last digit of the square to be $1$ or $5$.The cases where the last digit is $1$ are $x = 1$ and $x = 9$ (total $2$ cases).The case where the last digit is $5$ is $x = 5$ (total $1$ case).Total favorable outcomes = $2 + 1 = 3$.Total possible outcomes = $10$.Therefore,the required probability = $\frac{3}{10}$.

An integer is chosen at random and squared. The probability that the last digit of the square is $1$ or $5$ is

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