Two different numbers are randomly selected from the set ${ - 2, -1, 0, 3, 4, 5}$ and multiplied together. What is the probability that the product is $0$?
The same as the probability that either of them is zero.
The answer is 1/3. but I don't know how to get the answer.
P(1st=0) = 1/6
P(2nd=0) = 1/6
P(either) = 1/3
To find the probability that the product is $0$, we need to count how many combinations of two numbers from the given set result in a product of $0$, and then divide that count by the total number of possible combinations.
First, let's identify the pairs of numbers whose product is $0$. This occurs when at least one of the selected numbers is $0$. So, we can have the following combinations:
1. $0$ and $-2$: The product is $0 \times (-2) = 0$.
2. $0$ and $-1$: The product is $0 \times (-1) = 0$.
3. $0$ and $3$: The product is $0 \times 3 = 0$.
4. $0$ and $4$: The product is $0 \times 4 = 0$.
5. $0$ and $5$: The product is $0 \times 5 = 0$.
6. $-2$ and $0$: The product is $(-2) \times 0 = 0$.
7. $-1$ and $0$: The product is $(-1) \times 0 = 0$.
So, there are a total of $7$ combinations that result in a product of $0$.
Next, let's determine the total number of combinations that can be formed from the given set. Since we are selecting two numbers from a set of $6$ numbers, using the concept of combinations, the total number of combinations can be calculated as ${6 \choose 2} = \frac{6!}{2!(6-2)!} = 15$.
Finally, we can calculate the probability of the product being $0$ by dividing the number of combinations resulting in a product of $0$ by the total number of combinations:
Probability $= \frac{\text{Number of combinations resulting in a product of } 0}{\text{Total number of combinations}} = \frac{7}{15}.$
Therefore, the probability that the product is $0$ is $\frac{7}{15}$.