A bag contains 40 tickets numbered from 1 to 40. Two tickets are drawn from the bag without replacement. The probability that the `2^("nd")` ticket is a perfect square given that the `1^("st")` ticket was a perfect square is

Options

A. `(1)/(6)`
B. `(5)/(39)`
C. `(3)/(20)`
D. `(1)/(8)`

Answer

To solve the problem, we need to find the probability that the second ticket drawn is a perfect square given that the first ticket drawn was a perfect square. Let's break this down step by step.

### Step 1: Identify the Perfect Squares
First, we need to identify the perfect squares between 1 and 40. The perfect squares in this range are:
- \(1^2 = 1\)
- \(2^2 = 4\)
- \(3^2 = 9\)
- \(4^2 = 16\)
- \(5^2 = 25\)
- \(6^2 = 36\)

Thus, the perfect squares are: **1, 4, 9, 16, 25, 36**. This gives us a total of **6 perfect squares**.

### Step 2: Determine the Total Tickets
The total number of tickets in the bag is **40**.

### Step 3: Analyze the First Draw
We are given that the first ticket drawn is a perfect square. After drawing one perfect square ticket, we have:
- Remaining perfect squares: \(6 - 1 = 5\)
- Total remaining tickets: \(40 - 1 = 39\)

### Step 4: Calculate the Probability for the Second Ticket
Now, we need to find the probability that the second ticket drawn is also a perfect square. Given that there are **5 perfect squares left** and **39 tickets remaining**, the probability can be calculated as follows:

\[ P(\text{Second ticket is a perfect square | First ticket is a perfect square}) = \frac{\text{Number of remaining perfect squares}}{\text{Total remaining tickets}} = \frac{5}{39} \]

### Final Answer
Thus, the probability that the second ticket is a perfect square given that the first ticket was a perfect square is:

\[ \boxed{\frac{5}{39}} \]