Question

a bag contains 5 red, 4 green, and 3 blue marbles. what is the probability of randomly selecting a blue marble, replacing it in the bag, and then randomly selecting a red marble?
\\(\frac{1}{48}\\)
\\(\frac{1}{12}\\)
\\(\frac{5}{48}\\)
\\(\frac{5}{12}\\)

Explanation:

Step1: Calculate probability of blue marble

The total number of marbles is $5 + 4+3=12$. The probability of selecting a blue marble $P(B)$ is $\frac{3}{12}=\frac{1}{4}$ since there are 3 blue marbles out of 12.

Step2: Calculate probability of red marble

Since the marble is replaced, the total number of marbles is still 12. The probability of selecting a red marble $P(R)$ is $\frac{5}{12}$ as there are 5 red marbles out of 12.

Step3: Calculate combined probability

Since the two events are independent (because of replacement), the probability of both events occurring is the product of their individual probabilities. So $P = P(B)\times P(R)=\frac{1}{4}\times\frac{5}{12}=\frac{5}{48}$.

Answer:

$\frac{5}{48}$