Question

Two digits $a$ and $b$ are chosen at random without replacement. The probability that $a^2+b^2$ is divisible by $10$, is

• A. $\frac{3}{50}$
• B. $\frac{1}{{10}^2}$
• C. $\frac{4}{50}$
• D. $\frac{9}{50}$

Answer 1

Answer: (D)

$\frac{9}{50}$

We know that any number can be formed from $0,1,2,3,4,5,6,7,8,9$
Let $E$ be the desired event.
since we know that a number to be divisible by $10$ it must have $0$ in the end (unit place).
which means $a^2$ and $b^2$ should add up to make $10$.
This is possible only if the last digit of $a^2$ and $b^2$ have the combination of $(1,9)(4,6)$ and $(5,5)$
now we can see that $1$ comes at the unit place in $2$ cases ($1^2$ and $9^2$) similarly we can see that $9,4,6$ comes at the unit place in two cases and $5$ comes at unit place in one case i.e $5^2$
hence $P\left(E\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)\{P\left(1\right)P\left(9\right)+P\left(4\right)P\left(6\right)\}+\left(\begin{array}{c}2\\ 1\end{array}\right)\times P\left(5\right)P\left(5\right)=2\times \{\frac{2}{10}\times \frac{2}{10}+\frac{2}{10}\times \frac{2}{10}\}+2\times \frac{1}{10}\times \frac{1}{10}=\frac{9}{50}$