Question

A box contains 100 tickets numbered 00,01,02,....99 and a ticket is drawn at random. Let X denote the sum of the digits on that ticket and Y denote the product of those digits. Then the value of P(X=2IY=0) is

• A. \(\frac{3}{19}\)
• B. \(\frac{6}{19}\)
• C. \(\frac{1}{19}\)
• D. \(\frac{2}{19}\)
• E. \(\frac{1}{100}\)

Answer 1

The Correct Option is D

Given:

• A box contains 100 tickets numbered from 00 to 99
• Random variable X = sum of digits on the ticket
• Random variable Y = product of digits on the ticket

Step 1: Find all tickets where \( Y = 0 \) (product of digits is 0):

These are tickets where at least one digit is 0: \[ \{00, 01, 02, ..., 09, 10, 20, ..., 90\} \] Total of 19 such tickets (00-09 = 10, plus 10-90 in steps of 10 = 9, totaling 19).

Step 2: Find tickets where both \( Y = 0 \) and \( X = 2 \):

These are: \[ \{02, 11, 20\} \] But only 02 and 20 satisfy both conditions (11 has product 1 ≠ 0).

Step 3: Calculate the conditional probability: \[ P(X = 2 | Y = 0) = \frac{\text{Number of tickets with } X=2 \text{ and } Y=0}{\text{Number of tickets with } Y=0} = \frac{2}{19} \]

The correct answer is (D) \( \frac{2}{19} \).

Answer 2

Let \( X \) be the sum of the digits on the ticket and \( Y \) be the product of the digits. 

We want to find \( P(X=2 \mid Y=0) \). 

Using the definition of conditional probability:

\[ P(X=2 \mid Y=0) = \frac{P(X=2 \text{ and } Y=0)}{P(Y=0)} \]

First, let's find the number of tickets where \( Y=0 \). The product of the digits is 0 if at least one of the digits is 0.

• Tickets with \( Y=0 \): There are 19 such tickets: 00, 10, 20, ..., 90. (Note we're considering 00 as a valid ticket).

Now let's find the number of tickets where both \( X=2 \) and \( Y=0 \). This means the sum of digits is 2, and at least one digit is 0. The only possibilities are the tickets "20" and "02". There are 2 such tickets.

Therefore:

\[ P(X=2 \text{ and } Y=0) = \frac{2}{100} = \frac{1}{50} \] \[ P(Y=0) = \frac{19}{100} \] \[ P(X=2 \mid Y=0) = \frac{\frac{1}{50}}{\frac{19}{100}} = \frac{1}{50} \cdot \frac{100}{19} = \frac{2}{19} \]

Therefore, the value of \( P(X=2 \mid Y=0) \) is \( \frac{2}{19} \).