Two different numbers are chosen at random from the set {1, 2, 3, ..., 5}. What is the probability that the product of the numbers is even?

Answer 1

To determine the probability that the product of two randomly chosen numbers from the set $\{1,2,3,4,5\}$ is even, we follow these steps:

1. **Calculate the total number of possible pairs:**
Since we are selecting 2 different numbers from a set of 5, we use combinations:
$$(\genfrac{}{}{0ex}{}{5}{2})=\frac{5!}{2!(5-2)!}=\frac{5\times 4}{2\times 1}=10$$
There are 10 possible pairs.

2. **List all pairs:**
The pairs are: $(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5)$.

3. **Identify pairs with even products:**
A product is even if at least one of the numbers is even. The even numbers in the set are 2 and 4.
We count the pairs that include at least one even number.

- Pairs including number 2: $(1,2),(2,3),(2,4),(2,5)$
- Pairs including number 4 (excluding those already counted with 2): $(1,4),(3,4),(4,5)$

These are:
$(1,2)$, $(2,3)$, $(2,4)$, $(2,5)$, $(1,4)$, $(3,4)$, $(4,5)$

That's 7 pairs where the product is even.

4. **Calculate the probability:**
The probability is the number of favorable outcomes divided by the total number of outcomes.
$$P(\text{even product})=\frac{7}{10}$$

Thus, the probability that the product of the numbers is even is:
$$\boxed{{\displaystyle \frac{7}{10}}}$$

Answer 2

there are 5C2 = 10 pairs of numbers.
There are 3 odd numbers, and thus 3C2 = 3 pairs of odd numbers
That means that the product is even for the other 7 pairs, and so P(even) = 7/10

Answer 3

Your reasoning is correct, and you have succinctly summarized the solution to the problem. Let's walk through it step by step to confirm the details:

1. **Total number of pairs:**
$$(\genfrac{}{}{0ex}{}{5}{2})=\frac{5!}{2!(5-2)!}=\frac{5\times 4}{2\times 1}=10$$
This tells us there are 10 possible pairs of numbers that can be chosen.

2. **Counting pairs of odd numbers:**
The odd numbers in the set $\{1,2,3,4,5\}$ are \{1, 3, 5\}.
The number of ways to choose 2 numbers from these 3 odd numbers is:
$$(\genfrac{}{}{0ex}{}{3}{2})=\frac{3!}{2!(3-2)!}=\frac{3\times 2}{2\times 1}=3$$
So, there are 3 pairs of odd numbers: $(1,3),(1,5),(3,5)$.

3. **Counting pairs that result in an even product:**
Any number pair that includes at least one even number results in an even product. The complementary set of 10 total pairs minus the 3 pairs of odd numbers yields:
$$10-3=7$$
Thus, there are 7 pairs where the product is even.

4. **Calculating the probability:**
The probability is the number of pairs with an even product divided by the total number of possible pairs:
$$P(\text{even product})=\frac{7}{10}$$

Hence, the probability that the product of the numbers is even, using your method, is indeed:
$$\boxed{{\displaystyle \frac{7}{10}}}$$