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:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
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:
\[
\boxed{\frac{7}{10}}
\]
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?
3 answers
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
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
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:**
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
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:
\[
\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2}{2 \times 1} = 3
\]
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:
\[
\boxed{\frac{7}{10}}
\]
1. **Total number of pairs:**
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
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:
\[
\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2}{2 \times 1} = 3
\]
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:
\[
\boxed{\frac{7}{10}}
\]
0 answers