A bag contains 5 blue marbles, 3 red marbles, and 4 yellow marbles.

What is the probability of choosing one yellow marble and then a red marble with replacement?

A. [tex]$\frac{1}{12}$[/tex]
B. [tex]$\frac{7}{12}$[/tex]
C. [tex]$\frac{1}{11}$[/tex]
D. [tex]$\frac{5}{36}$[/tex]



Answer :

To determine the probability of choosing a yellow marble and then a red marble with replacement from a bag containing 5 blue marbles, 3 red marbles, and 4 yellow marbles, we can follow these steps:

1. Calculate the total number of marbles in the bag:
[tex]\[ \text{Total marbles} = 5 (\text{blue}) + 3 (\text{red}) + 4 (\text{yellow}) = 12 \text{ marbles} \][/tex]

2. Find the probability of choosing a yellow marble first:
The number of yellow marbles in the bag is 4.
[tex]\[ \text{Probability of picking a yellow marble} = \frac{4}{12} = \frac{1}{3} \approx 0.33333 \][/tex]

3. Find the probability of choosing a red marble next:
Since we are choosing with replacement, after choosing the first marble and replacing it, the total number of marbles remains 12. The number of red marbles in the bag is 3.
[tex]\[ \text{Probability of picking a red marble} = \frac{3}{12} = \frac{1}{4} = 0.25 \][/tex]

4. Calculate the combined probability of both events occurring:
These events are independent because the first marble is replaced back into the bag. Therefore, the probability of both events happening is the product of their individual probabilities.
[tex]\[ \text{Combined probability} = \left(\frac{1}{3}\right) \times \left(\frac{1}{4}\right) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \approx 0.08333 \][/tex]

Therefore, the probability of choosing one yellow marble and then a red marble with replacement is:
[tex]\[ \boxed{\frac{1}{12}} \][/tex]