Question

a bag contains eight yellow marbles, nine green marbles, three purple marbles, and five red marbles. two marbles are chosen from the bag. what expression would give the probability that one marble is yellow and the other marble is red?
\\( p(y \text{ and } r) = \frac{(_{8}p_{1})(_{5}p_{1})}{_{25}p_{2}} \\)
\\( p(y \text{ and } r) = \frac{(_{8}c_{1})(_{5}c_{1})}{_{25}c_{2}} \\)
\\( p(y \text{ and } r) = \frac{(_{1}c_{8})(_{1}c_{5})}{_{2}c_{25}} \\)
\\( p(y \text{ and } r) = \frac{(_{1}p_{8})(_{1}p_{5})}{_{2}p_{25}} \\)

Explanation:

Step1: Determine Total Marbles

First, calculate the total number of marbles. Yellow: 8, Green: 9, Purple: 3, Red: 5. Total = \(8 + 9 + 3 + 5 = 25\) marbles. We are choosing 2 marbles, and the order of choosing (yellow then red or red then yellow) doesn't matter here (since we just want one yellow and one red, regardless of order), so we use combinations (not permutations).

Step2: Favorable Outcomes

The number of ways to choose 1 yellow marble from 8 is \(_{8}C_{1}\), and the number of ways to choose 1 red marble from 5 is \(_{5}C_{1}\). The total number of ways to choose 2 marbles from 25 is \(_{25}C_{2}\).

Step3: Probability Formula

Probability is (number of favorable outcomes) / (total number of possible outcomes). So the probability \(P(Y \text{ and } R)\) is \(\frac{(_{8}C_{1})(_{5}C_{1})}{_{25}C_{2}}\).

Answer:

\( P(Y \text{ and } R) = \frac{(_{8}C_{1})(_{5}C_{1})}{_{25}C_{2}} \) (the second option)