Probability Analysis of Even Sums Greater than 5 from a Random Integer Selection
Understanding the probability that the sum of two randomly selected integers, with replacement, from the integers 1 to 10 is even and greater than 5 involves a step-by-step breakdown of the problem.
Determining Total Outcomes
When selecting two integers from 1 to 10 with replacement, the total number of possible outcomes is:
10 times 10 100
Conditions for the Sum
The sum of two integers can be even if:
Both integers are even. Both integers are odd.The even integers from 1 to 10 are: 2, 4, 6, 8, 10, totaling 5 even integers.
The odd integers from 1 to 10 are: 1, 3, 5, 7, 9, also totaling 5 odd integers.
Calculating Favorable Outcomes
We need to count the outcomes where the sum is even and greater than 5. Let's evaluate each case:
Case 1: Both Integers are Even
Possible even pairs are: (2,2), (2,4), (2,6), (2,8), (2,10), (4,2), (4,4), (4,6), (4,8), (4,10), (6,2), (6,4), (6,6), (6,8), (6,10), (8,2), (8,4), (8,6), (8,8), (8,10), (10,2), (10,4), (10,6), (10,8), (10,10).
The sums of these pairs are as follows:
2 2 4 (not counted) 2 4 6 (counted) 2 6 8 (counted) 2 8 10 (counted) 2 10 12 (counted) 4 2 6 (counted) 4 4 8 (counted) 4 6 10 (counted) 4 8 12 (counted) 4 10 14 (counted)And so on. The valid even pairs sums greater than 5 are:
(2,4), (2,6), (2,8), (2,10), (4,2), (4,4), (4,6), (4,8), (4,10), (6,2), (6,4), (6,6), (6,8), (6,10), (8,2), (8,4), (8,6), (8,8), (8,10), (10,2), (10,4), (10,6), (10,8), (10,10)Total valid pairs sums greater than 5: 20
Case 2: Both Integers are Odd
Possible odd pairs are: (1,1), (1,3), (1,5), (1,7), (1,9), (3,1), (3,3), (3,5), (3,7), (3,9), (5,1), (5,3), (5,5), (5,7), (5,9), (7,1), (7,3), (7,5), (7,7), (7,9), (9,1), (9,3), (9,5), (9,7), (9,9).
The sums of these pairs are as follows:
1 1 2 (not counted) 1 3 4 (not counted) 1 5 6 (counted) 1 7 8 (counted) 1 9 10 (counted) 3 1 4 (not counted) 3 3 6 (counted) 3 5 8 (counted) 3 7 10 (counted) 3 9 12 (counted)And so on. The valid odd pairs sums greater than 5 are:
(1,5), (1,7), (1,9), (3,3), (3,5), (3,7), (3,9), (5,1), (5,3), (5,5), (5,7), (5,9), (7,1), (7,3), (7,5), (7,7), (7,9), (9,1), (9,3), (9,5), (9,7), (9,9)Total valid pairs sums greater than 5: 20
Total Valid Outcomes
The total number of favorable outcomes for both cases is:
20 (even) 20 (odd) 40
Calculating Probability
The probability that the sum of the two selected integers is both even and greater than 5 is:
P_{even and > 5} frac{text{Number of valid outcomes}}{text{Total outcomes}} frac{40}{100} frac{2}{5}
Final Answer:
The probability that the sum of the two numbers is even and greater than 5 is: boxed{frac{2}{5}}