Question

Two natural numbers are selected at random, find the probability that their sum is divisible by $10$.

• A. $\frac{4}{15}$
• B. $\frac{2}{5}$
• C. $\frac{1}{10}$
• D. $\frac{3}{10}$

Answer 1

The correct option is C $\frac{1}{10}$

Let the two non-negative integers selected be $a$ and $b$.
Then $a=10\alpha +a_1$ and $b=10\beta +b_1$,
where $\alpha ,\beta \ge 0$ and $0<a_1,b_1\le 9$
Thus, number of possible cases for $(a_1,b_1)$ is $10\times 10=100$
Out of these $100$, $(0,0),(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2)$ and $(9,1)$ are favourable.
Hence, the probability of the required event is $\frac{10}{100}=\frac{1}{10}$