Question

two integers are randomly selected from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and are added together. find the probability that their sum is 5 if they are selected (a) with replacement, (b) without replacement. (a) the probability that their sum is 5 if they are selected with replacement is . (type an integer or a simplified fraction.)

Explanation:

Step1: Calculate total outcomes with replacement

The set has 9 elements. When selecting two integers with replacement, the total number of outcomes is \(n = 9\times9=81\) since for the first - number there are 9 choices and for the second number there are also 9 choices.

Step2: Find pairs that sum to 5 with replacement

The pairs \((x,y)\) such that \(x + y=5\) and \(x,y\in\{1,2,3,4,5,6,7,8,9\}\) are \((1,4)\), \((2,3)\), \((3,2)\), \((4,1)\). So there are \(m = 4\) favorable outcomes.

Step3: Calculate probability with replacement

The probability \(P=\frac{m}{n}=\frac{4}{81}\).

Step4: Calculate total outcomes without replacement

When selecting two integers without replacement, the total number of outcomes is \(n=\text{C}_{9}^2=\frac{9!}{2!(9 - 2)!}=\frac{9\times8}{2\times1}=36\).

Step5: Find pairs that sum to 5 without replacement

The pairs \((x,y)\) such that \(x + y = 5\) and \(x,y\in\{1,2,3,4,5,6,7,8,9\}\) without replacement are \((1,4)\), \((2,3)\). So there are \(m = 2\) favorable outcomes.

Step6: Calculate probability without replacement

The probability \(P=\frac{m}{n}=\frac{2}{36}=\frac{1}{18}\).

Answer:

(a) \(\frac{4}{81}\)
(b) \(\frac{1}{18}\)