A box contains 4 tickets numbered 1, 2, 3, and 4. Two tickets are drawn one after the other without replacement. Find the probability that their sum is a prime number.

Since two tickets are drawn one after the other without replacement, the order matters.
The first ticket can be any of the 4 numbers.
The second ticket can be any of the remaining 3 numbers.
Total number of possible outcomes = $4 \times 3 = 12$.
These outcomes are:
(1,2), (1,3), (1,4)
(2,1), (2,3), (2,4)
(3,1), (3,2), (3,4)
(4,1), (4,2), (4,3)

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The prime numbers relevant here are 2, 3, 5, 7, etc.

Let's list the sums:

• (1,2) Sum = $1+2=3$ (Prime)
• (1,3) Sum = $1+3=4$
• (1,4) Sum = $1+4=5$ (Prime)
• (2,1) Sum = $2+1=3$ (Prime)
• (2,3) Sum = $2+3=5$ (Prime)
• (2,4) Sum = $2+4=6$
• (3,1) Sum = $3+1=4$
• (3,2) Sum = $3+2=5$ (Prime)
• (3,4) Sum = $3+4=7$ (Prime)
• (4,1) Sum = $4+1=5$ (Prime)
• (4,2) Sum = $4+2=6$
• (4,3) Sum = $4+3=7$ (Prime)

The pairs with prime sums are:
(1,2), (1,4), (2,1), (2,3), (3,2), (3,4), (4,1), (4,3)
There are 8 favorable outcomes.

Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Simplify the fraction:

The probability that the sum is prime is $\frac{2}{3}$.