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(B) The last digit of a square depends only on the last digit of the integer.Let the last digit of the integer be $x \in \{0, 1, 2, 3, 4, 5, 6, 7, 8,

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$\frac{3}{10}$

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(B) The last digit of a square depends only on the last digit of the integer.Let the last digit of the integer be $x \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.The squares of these digits end in:$0^2 	o 0$$1^2 	o 1$$2^2 	o 4$$3^2 	o 9$$4^2 	o 6$$5^2 	o 5$$6^2 	o 6$$7^2 	o 9$$8^2 	o 4$$9^2 	o 1$The last digit of the square is $1$ if the integer ends in $1$ or $9$.The last digit of the square is $5$ if the integer ends in $5$.Thus,there are $3$ favorable outcomes out of $10$ possible digits.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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