Two whole numbers are randomly selected and multiplied. Consider two events `E_1 and E_2` defined as `E_1` : Their product is divisible by 5 and `E_2` Unit's place in their product is 5 Which of the following statement(s) is/are correct?

Options

A. `E_(1)` is twice as likely to occur as `E_(2)`
B. `E_(1)and E_(2)` are disjoint
C. `P(E_(2)//E^(1))=1//4`
D. `P(E_(1)//E_(2))=1`

Explanation

To solve the problem, we need to analyze the two events \( E_1 \) and \( E_2 \):

1. **Event \( E_1 \)**: The product of the two selected numbers is divisible by 5.
2. **Event \( E_2 \)**: The unit's place of the product is 5.

### Step 1: Calculate the probability of \( E_1 \)

To find the probability of \( E_1 \), we need to determine the conditions under which the product of two numbers is divisible by 5.

- A product is divisible by 5 if at least one of the numbers is divisible by 5.
- The possible unit digits of whole numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
- Among these, the digits that contribute to divisibility by 5 are 0 and 5.

The probability that a randomly selected whole number does not end with 0 or 5 (i.e., ends with 1, 2, 3, 4, 6, 7, 8, or 9) is \( \frac{8}{10} = \frac{4}{5} \).

Thus, the probability that both selected numbers do not end with 0 or 5 is:
\[ P(\text{both not ending with 0 or 5}) = \left(\frac{4}{5}\right) \times \left(\frac{4}{5}\right) = \frac{16}{25} \]

Therefore, the probability that at least one of them ends with 0 or 5 (which means the product is divisible by 5) is:
\[ P(E_1) = 1 - P(\text{both not ending with 0 or 5}) = 1 - \frac{16}{25} = \frac{9}{25} \]

### Step 2: Calculate the probability of \( E_2 \)

Next, we calculate the probability of \( E_2 \). The unit's place of the product of two numbers will be 5 if:
- One number ends with 5 and the other ends with an odd digit (1, 3, 5, 7, 9).

The probability that a randomly selected number ends with 5 is \( \frac{1}{10} \), and the probability that it ends with an odd digit (1, 3, 5, 7, 9) is \( \frac{5}{10} = \frac{1}{2} \).

Thus, the probability that one number ends with 5 and the other ends with an odd digit can be calculated as:
\[ P(E_2) = P(\text{first number ends with 5}) \times P(\text{second number ends with odd digit}) + P(\text{first number ends with odd digit}) \times P(\text{second number ends with 5}) \]
This gives us:
\[ P(E_2) = \left(\frac{1}{10} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{10}\right) = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \]

### Step 3: Calculate the conditional probability \( P(E_2 | E_1) \)

Now we need to find the conditional probability \( P(E_2 | E_1) \):
\[ P(E_2 | E_1) = \frac{P(E_2 \cap E_1)}{P(E_1)} \]

Since \( E_2 \) (unit's place is 5) implies \( E_1 \) (product is divisible by 5), we can say:
\[ P(E_2 \cap E_1) = P(E_2) \]

Thus:
\[ P(E_2 | E_1) = \frac{P(E_2)}{P(E_1)} = \frac{\frac{1}{10}}{\frac{9}{25}} = \frac{1}{10} \times \frac{25}{9} = \frac{25}{90} = \frac{5}{18} \]

### Conclusion

Based on the calculations:
- \( P(E_1) = \frac{9}{25} \)
- \( P(E_2) = \frac{1}{10} \)
- \( P(E_2 | E_1) = \frac{5}{18} \)