One ticket is selected at random from $$50$$ tickets numbered $$00, 01, 02, \ldots, 49$$. 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

Let the two events be defined as:
  • $$A$$ : “sum of the two digits is $$8$$”
  • $$B$$ : “product of the two digits is $$0$$” (i.e. at least one digit is zero)

We are asked to find the conditional probability $$P(A\mid B)$$. For equally likely outcomes,

$$P(A\mid B)=\frac{\text{number of tickets satisfying }A\text{ and }B}{\text{number of tickets satisfying }B}$$

Step 1 : Count the tickets that satisfy $$B$$
The admissible tickets are the two-digit strings from $$00$$ to $$49$$ (inclusive). A ticket has product zero if at least one digit is zero.

• Tens digit is zero (units digit can be $$0$$-$$9$$):
  $$00,01,02,03,04,05,06,07,08,09$$  ⇒  $$10$$ tickets.

• Units digit is zero while tens digit is $$1$$-$$4$$:
  $$10,20,30,40$$  ⇒  $$4$$ tickets.

There is no overlap except $$00$$, which is already in the first list, so

$$|B| = 10 + 4 = 14$$ tickets.

Step 2 : Among these, find those that also satisfy $$A$$
List the 14 tickets and compute each digit-sum:

$$\begin{array}{cccccccccc} 00 & 01 & 02 & 03 & 04 & 05 & 06 & 07 & 08 & 09 & 10 & 20 & 30 & 40 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 \end{array}$$

The only ticket with digit-sum $$8$$ is $$08$$. Hence

$$|A\cap B| = 1.$$

Step 3 : Compute the conditional probability

$$P(A\mid B)=\frac{|A\cap B|}{|B|}=\frac{1}{14}.$$

Therefore, the required probability is $$\frac{1}{14}$$.

Option A which is: $$\frac{1}{14}$$