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.

Here's the solution for Example 12: A box contains 4 tickets numbered 1, 2, 3, and 4. Two tickets are drawn one after the other without replacement. We need to find the probability that their sum is a prime number. Step 1: Determine the total number of possible outcomes. 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 × 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) Step 2: Calculate the sum for each pair and identify prime sums. 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) Step 3: Count the number of favorable outcomes (sums that are 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. Step 4: Calculate the probability. Probability = Number of favorable outcomesTotal number of possible outcomes P(sum is prime) = (8)/(12) Simplify the fraction: P(sum is prime) = (2)/(3) The probability that the sum is prime is (2)/(3). That's 2 down. 3 left today — send the next one.