Question

A two-digit number is formed with digits 2, 3, 5, 7, 9 without repetition. Find the probability of the following events:
Event A: The number formed is an odd number.
Event B: The number formed is a multiple of 5.

Hint:

When finding probabilities from digit-based combinations, always count possible outcomes based on digit placement (tens and units separately).

Answer 1

Step 1: Total possible two-digit numbers.
We have 5 digits: 2, 3, 5, 7, 9. To form a two-digit number without repetition: \[ \text{Total numbers} = 5 \times 4 = 20 \] Step 2: Event A — The number is odd.
For a number to be odd, the unit place must be an odd digit. Odd digits available = 3, 5, 7, 9 (4 choices). For each, tens place can be any of the remaining 4 digits. \[ \text{Total odd numbers} = 4 \times 4 = 16 \] \[ P(A) = \frac{16}{20} = \frac{4}{5} \] \[ \boxed{P(A) = \frac{4}{5}} \] Step 3: Event B — The number is a multiple of 5.
For a number to be a multiple of 5, the unit digit must be 5. Thus, unit place = 5 (1 choice). Tens place can be filled by any of the remaining 4 digits. \[ \text{Total numbers divisible by 5} = 4 \] \[ P(B) = \frac{4}{20} = \frac{1}{5} \] \[ \boxed{P(B) = \frac{1}{5}} \] Step 4: Conclusion.
\[ P(A) = \frac{4}{5}, \quad P(B) = \frac{1}{5} \]