To solve the problem, we need to calculate the probability that the first ball drawn is red and the second ball drawn is blue, without replacement. Let's proceed step by step.
1. Determine the total number of balls in the bag:
- Red balls: 4
- Blue balls: 3
- Green balls: 5
Total number of balls is:
[tex]\( 4 + 3 + 5 = 12 \)[/tex]
2. Calculate the probability that the first ball drawn is red:
There are 4 red balls out of a total of 12 balls.
[tex]\[
\text{Probability that the first ball is red} = \frac{4}{12} = \frac{1}{3} \approx 0.3333
\][/tex]
3. Calculate the probability that the second ball drawn is blue given the first ball was red:
If the first ball drawn is red, there are now 11 balls remaining (since we are drawing without replacement).
Initially, there are 3 blue balls in the bag, and the number of blue balls remains unchanged (3) since we drew a red ball first.
The probability that the second ball is blue given that the first ball was red is:
[tex]\[
\text{Probability that the second ball is blue} = \frac{3}{11} \approx 0.2727
\][/tex]
4. Calculate the combined probability of both events happening:
These two events are dependent, so we multiply their probabilities:
[tex]\[
\text{Total probability} = \left( \frac{1}{3} \right) \times \left( \frac{3}{11} \right) = \frac{1 \times 3}{3 \times 11} = \frac{3}{33} = \frac{1}{11} \approx 0.0909
\][/tex]
Therefore, the probability that the first ball is red and the second ball is blue is [tex]\( \frac{1}{11} \)[/tex].
The answer is:
C. [tex]\( \frac{1}{11} \)[/tex]
