Probability: Harder Probability - Using Combinations

>$$S = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$$ If two different numbers are randomly chosen at the same time from set $$S$$, then what is the probability that their product is divisible by 6?

Next, find the total number of choices. Since you're choosing two numbers from a set of nine numbers, and the order doesn't matter, the total number of possible combinations is >$$ C(9,2) = \frac{9 \times 8}{2} = 36$$ Finally, to answer the question, divide the number of ways of getting a product that is divisible by 6 by the total number of choices and simplify the result. >$$\text{Probability} = \frac{14}{36} = \frac{7}{18}$$

Incorrect. It's possible to arrive at $$\frac{29}{81}$$ if you incorrectly calculated the total number of choices. Notice that the problem states that the two numbers must be different. [[snippet]]

Incorrect. An answer of $$\frac{1}{3}$$ suggests you might not have realized that if one of the numbers is 6, then the other number can be anything, and the product will be divisible by 6. [[snippet]]

Incorrect. It's possible to get $$ \frac{1}{6}$$ if you didn't consider the case in which one of the numbers is 6. [[snippet]]

Incorrect. You may have gotten $$ \frac{2}{9}$$ if you didn't realize that since $$6 = 2 \cdot 3$$, you can get the product you're looking for by multiplying any even number by any number that's divisible by 3. [[snippet]]

Right! To find the probability, divide the number of choices that make the product divisible by 6 by the total number of choices. > $$\displaystyle \text{Probability}= \frac{\text{Number of Wanted Combinations}}{\text{Total Number of Possible Combinations}}$$ In this problem, you're choosing the two numbers at the same time, so order doesn't matter, and the two numbers must be different. First, find the number of choices that make the product divisible by 6. There are two ways to do this: In Case 1, one of the numbers is 6. In this case, the other number can be anything. Therefore, one of the numbers must be 6, and the other can be any of the remaining eight numbers in $$S$$, so there are $$1 \times 8 = 8 $$ possible combinations. In Case 2, since $$6 = 2 \times 3$$, the product will be divisible by 6 if one of the numbers is even and the other is divisible by 3. In this case, you should ignore the 6 in the set since you already dealt with that in Case 1. Besides 6, there are three even numbers in $$S$$ (2, 4, and 8) and two numbers that are divisible by 3 (3 and 9). So you have to choose one of three numbers for the first value and one of two numbers for the second, for a total of $$3 \times 2 = 6$$ possible combinations. Combining Cases 1 + 2, thus, the total is $$8+6 = 14$$ possible combinations.

$$\frac{1}{6}$$

$$\frac{2}{9}$$

$$\frac{1}{3}$$

$$\frac{29}{81}$$

$$\frac{7}{18}$$

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