Probability of Summing Even Numbers from Tickets with Replacement

When selecting three tickets at random with replacement from a bag containing tickets numbered from 1 to 100, the question arises: What is the probability that the sum of the numbers on the tickets is an even number?

Let's delve into the analysis step by step to determine this probability.

Step 1: Determining the Parity of the Numbers

We start by understanding the even and odd numbers within the range from 1 to 100.

Even numbers: 2, 4, 6, ..., 100 (total 50 numbers) Odd numbers: 1, 3, 5, ..., 99 (total 50 numbers)

Step 2: Understanding How the Sum Can Be Even

The sum of three numbers will be even if:

All three numbers are even. One number is odd and the other two are even.

Step 3: Calculating the Probabilities for Each Case

Case 1: All Three Numbers are Even

The probability of selecting an even number on a single draw is:

1 / 2 50/100

The probability of selecting three even numbers is:

(1/2)3 1/8

Case 2: One Odd and Two Even

The probability of selecting an odd number on a single draw is also:

1 / 2 50/100

The probability of selecting two even numbers is:

(1/2)2 1/4

The total combinations of selecting one odd and two even numbers can occur in three different orders (OEE, EOE, EEO), each with the same probability:

3 * (1/2) * (1/4) 3/8

Step 4: Combining the Probabilities

We now sum the probabilities from both cases to get the total probability that the sum is even:

P(even sum) P(all even) P(one odd and two even) 1/8 3/8 4/8 1/2

Final Result: The probability that the sum of the three numbers drawn is an even number is:

1/2

Note

This solution assumes that the drawings are with replacement, meaning each draw does not affect the others. If the drawings were without replacement, the problem would require a different approach due to the changing probabilities with each draw. This scenario helps in understanding the correct application of probability principles when dealing with random selections.