A bag contains 63 cards (numbered 1,2,3,... 63). Two cards are picked at random from the bag (one after another and without replacement). What is the probability that the sum of the numbers of both the cards drawn is even?

Options

A. `34/63`
B. `7/11`
C. `11/63`
D. Other than those given as options

Answer

To solve the problem, we need to find the probability that the sum of the numbers on two cards drawn from a bag of 63 cards is even.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 63 cards numbered from 1 to 63.
- We will draw two cards one after another without replacement.
- We need to find the probability that the sum of the numbers on the two drawn cards is even.

2. **Identifying Conditions for Even Sum**:
- A sum is even if both numbers are even or both numbers are odd.
- Therefore, we need to consider two cases:
- Case 1: Both cards drawn are even.
- Case 2: Both cards drawn are odd.

3. **Counting Even and Odd Cards**:
- The even numbers from 1 to 63 are: 2, 4, 6, ..., 62.
- There are 31 even numbers.
- The odd numbers from 1 to 63 are: 1, 3, 5, ..., 63.
- There are 32 odd numbers.

4. **Calculating Probability for Case 1 (Both Cards Even)**:
- The probability of drawing the first even card:
\[ P(\text{first even}) = \frac{31}{63} \]
- After drawing one even card, there are now 30 even cards left out of a total of 62 cards:
\[ P(\text{second even | first even}) = \frac{30}{62} \]
- Therefore, the probability for Case 1 is:
\[ P(\text{both even}) = P(\text{first even}) \times P(\text{second even | first even}) = \frac{31}{63} \times \frac{30}{62} \]

5. **Calculating Probability for Case 2 (Both Cards Odd)**:
- The probability of drawing the first odd card:
\[ P(\text{first odd}) = \frac{32}{63} \]
- After drawing one odd card, there are now 31 odd cards left out of a total of 62 cards:
\[ P(\text{second odd | first odd}) = \frac{31}{62} \]
- Therefore, the probability for Case 2 is:
\[ P(\text{both odd}) = P(\text{first odd}) \times P(\text{second odd | first odd}) = \frac{32}{63} \times \frac{31}{62} \]

6. **Total Probability**:
- The total probability that the sum is even is the sum of the probabilities of both cases:
\[ P(\text{even sum}) = P(\text{both even}) + P(\text{both odd}) \]
- Substituting the values:
\[ P(\text{even sum}) = \left(\frac{31}{63} \times \frac{30}{62}\right) + \left(\frac{32}{63} \times \frac{31}{62}\right) \]

7. **Simplifying the Expression**:
- Factor out the common terms:
\[ P(\text{even sum}) = \frac{31}{63} \times \left(\frac{30}{62} + \frac{32}{62}\right) \]
- Simplifying further:
\[ P(\text{even sum}) = \frac{31}{63} \times \frac{62}{62} = \frac{31}{63} \]

### Final Answer:
The probability that the sum of the numbers on both cards drawn is even is:
\[ \frac{31}{63} \]