Question

One ticket is selected at random from $50$ tickets numbered $00,01,\mathrm{02....49}$. Then the probability that the sum of the digits on the selected ticket is $8$, given that the product of these digits is zero, is

• A. $\frac{1}{50}$
• B. $\frac{1}{25}$
• C. $\frac{1}{100}$
• D. $\frac{49}{50}$

Answer 1

Answer: (A)

$\frac{1}{50}$

$S=7$ and $T=6$ can happen if the digits are: $1$ and $6$. There can thus be $2$ numbers $16$ and $61$.
The probability $=\frac{2}{100}=\frac{1}{50}$