The lowest 10 prime numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.

We need to find the probability that the sum of two different prime numbers chosen from this list ends with the digit 2.

To find this, we first identify all possible pairs of these prime numbers and their corresponding sums. Then, we count the pairs where the sum ends with 2.

Step 1: List all pairs and their sums

For two different prime numbers $p$ and $q$ where $p < q$:

• (2, 3) → 2 + 3 = 5

• (2, 5) → 2 + 5 = 7

• (2, 7) → 2 + 7 = 9

• (2, 11) → 2 + 11 = 13

• (2, 13) → 2 + 13 = 15

• (2, 17) → 2 + 17 = 19

• (2, 19) → 2 + 19 = 21

• (2, 23) → 2 + 23 = 25

• (2, 29) → 2 + 29 = 31

• (3, 5) → 3 + 5 = 8

• (3, 7) → 3 + 7 = 10

• (3, 11) → 3 + 11 = 14

• (3, 13) → 3 + 13 = 16

• (3, 17) → 3 + 17 = 20

• (3, 19) → 3 + 19 = 22

• (3, 23) → 3 + 23 = 26

• (3, 29) → 3 + 29 = 32

• (5, 7) → 5 + 7 = 12

• (5, 11) → 5 + 11 = 16

• (5, 13) → 5 + 13 = 18

• (5, 17) → 5 + 17 = 22

• (5, 19) → 5 + 19 = 24

• (5, 23) → 5 + 23 = 28

• (5, 29) → 5 + 29 = 34

• (7, 11) → 7 + 11 = 18

• (7, 13) → 7 + 13 = 20

• (7, 17) → 7 + 17 = 24

• (7, 19) → 7 + 19 = 26

• (7, 23) → 7 + 23 = 30

• (7, 29) → 7 + 29 = 36

• (11, 13) → 11 + 13 = 24

• (11, 17) → 11 + 17 = 28

• (11, 19) → 11 + 19 = 30

• (11, 23) → 11 + 23 = 34

• (11, 29) → 11 + 29 = 40

• (13, 17) → 13 + 17 = 30

• (13, 19) → 13 + 19 = 32

• (13, 23) → 13 + 23 = 36

• (13, 29) → 13 + 29 = 42

• (17, 19) → 17 + 19 = 36

• (17, 23) → 17 + 23 = 40

• (17, 29) → 17 + 29 = 46

• (19, 23) → 19 + 23 = 42

• (19, 29) → 19 + 29 = 48

• (23, 29) → 23 + 29 = 52

Step 2: Count pairs where the sum ends with 2

The pairs whose sum ends with 2 are:

• (2, 19) → 21
• (3, 19) → 22
• (5, 7) → 12
• (5, 17) → 22
• (13, 19) → 32
• (19, 23) → 42
• (23, 29) → 52

There are 7 such pairs.

Step 3: Calculate the total number of pairs

The total number of ways to choose 2 different primes from 10 primes is: $\binom{10}{2} = 45$

Step 4: Find the probability

The probability that the sum of two randomly chosen different prime numbers ends with the digit 2 is: $\frac{7}{45}$

This fraction can be simplified: $\frac{7}{45} = \frac{7}{45} = \frac{7}{45}$

So, the final answer is: $\frac{7}{45}$

Would you like more details or have any questions?

Here are some related questions you might consider:

1. What is the probability that the sum of two different primes ends with an odd digit?
2. How many ways can you choose 3 different primes from the first 10 primes?
3. What is the sum of the first 10 prime numbers?
4. What is the probability that the sum of two different primes ends with the digit 3?
5. How many pairs of the first 10 prime numbers have a sum that is also a prime number?
6. What is the sum of the largest and smallest prime numbers in the first 10 primes?
7. How can we determine if a number is a prime?
8. What are the properties of prime numbers?

Tip: When working with probabilities, ensure the total number of outcomes is accurately counted to avoid errors in calculation.