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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?

[tex]\[ P(Y \text{ and } R) = \frac{C_8^1 \cdot C_5^1}{C_{25}^2} \][/tex]



Answer :

GinnyAnswer
Let's solve the problem step by step:

1. Count the total number of marbles in the bag:
[tex]\[ \text{Total number of marbles} = 8 + 9 + 3 + 5 = 25 \][/tex]

2. Calculate the number of ways to choose 1 yellow marble from 8 yellow marbles:
[tex]\[ \binom{8}{1} = 8 \][/tex]

3. Calculate the number of ways to choose 1 red marble from 5 red marbles:
[tex]\[ \binom{5}{1} = 5 \][/tex]

4. Calculate the total number of ways to choose 2 marbles out of the 25 total marbles:
[tex]\[ \binom{25}{2} = \frac{25 \times 24}{2 \times 1} = 300 \][/tex]

5. Put it all together to find the probability that one marble is yellow and the other marble is red. The expression for this probability is given by:
[tex]\[ P(Y \text{ and } R) = \frac{\binom{8}{1} \times \binom{5}{1}}{\binom{25}{2}} \][/tex]

6. Substitute the values we calculated:
[tex]\[ P(Y \text{ and } R) = \frac{8 \times 5}{300} \][/tex]

7. Simplify the fraction:
[tex]\[ P(Y \text{ and } R) = \frac{40}{300} = \frac{2}{15} \approx 0.1333 \][/tex]

Therefore, the probability that one marble is yellow and the other marble is red is approximately [tex]\(0.1333\)[/tex] or [tex]\(\frac{2}{15}\)[/tex]. The correct expression from the given options would correspond to:

[tex]\[ P(Y \text{ and } R) = \frac{\left(C_8 C_1\right)\left(C_5 C_1\right)}{\binom{25}{2}} \][/tex]

where [tex]\( C_8C_1 = \binom{8}{1} \)[/tex] and [tex]\( C_5C_1 = \binom{5}{1} \)[/tex].

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