Question

Two numbers a and b are selected from the set of natural number then the probability that $a^2+b^2$ is divisible by $5$ is?

• A. $\frac{9}{25}$
• B. $\frac{7}{18}$
• C. $\frac{11}{36}$
• D. $\frac{17}{81}$

Answer 1

The correct option is A $\frac{9}{25}$

$a$ and $b$ are chosen from natural numbers.

$N=\{1,2,3,4,.......\}$

Let $p$ be a natural number.

then $p$ is of the form $5k$ or $\begin{array}{c}5k+1\\ 5k-1\end{array}$ or $\begin{array}{c}5k+2\\ 5k-2\end{array}$

$p^2=\{\begin{array}{c}25K^2\\ 25K^2\pm 10K+1\\ 25K^2\pm 20K+4\end{array}$

So, The remainder of $\frac{P^2}{5}$ would be either $0$ or $1$ or $-1$.

Case (i): Both the numbers $a$ & $b$ come from the set whose square gives reminder $0$ with $5$. Number of possibilities $=\frac{1}{5}\times \frac{1}{5}=\frac{1}{25}$

Case (i|): One no. '$a$' gives reminder '$1$' with $5$ and '$b$' gives '$-1$' and vice versa.

So, Number of ways $={}^2{C}_1.\left(\frac{2}{5}\right)\left(\frac{2}{5}\right)$

$=\frac{8}{25}$.

$\therefore$ Required probability $=\frac{9}{25}$