Question

One ticket is selected at random from$50$tickets numbered$00,01,02,03,...................,49$. Then the probability that the sum of the digits on the selected ticket is$8$, given that the product of these digits is zero, equals

Answer 1

Any number in the set S={$00,01,02,03,...................,49$} is of the form$\text{ab}$.

where$a\in \{0,1,2,3,4\}andb\in \{0,1,2,3.....,9\}$

For the product of digits to be zero, the number must be of the form either$x0$, which are$5$in numbers, because$x\in \{0,1,2,3,4\}$

The form$0x$, which are$10$in numbers because$x\in \{0,1,2,3.........,9\}$

The only number common to both is$=00$.

Thus the number of numbers in$S$, the product of whose digits is zero$=10+5-1=14$.

The number whose sum of digits is$8$is just one,$i.e.08$

The required probability$=\frac{1}{14}$.