Certainly! Let's solve this step-by-step:
### Step 1: Calculate [tex]\( P(B) \)[/tex]
To find [tex]\( P(B) \)[/tex], the probability of choosing a black ball first:
- The total number of balls is [tex]\( 4 + 8 = 12 \)[/tex].
- The number of black balls is [tex]\( 8 \)[/tex].
Hence, the probability of choosing a black ball first is:
[tex]\[ P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{8}{12} = \frac{2}{3} \approx 0.6667 \][/tex]
### Step 2: Calculate [tex]\( P(R \mid B) \)[/tex]
To find [tex]\( P(R \mid B) \)[/tex], the probability of choosing a red ball second given that a black ball was chosen first:
- After choosing one black ball, we have [tex]\( 12 - 1 = 11 \)[/tex] balls remaining.
- The number of red balls remains [tex]\( 4 \)[/tex] since none have been chosen yet.
Thus, the probability of choosing a red ball second given that a black ball was chosen first is:
[tex]\[ P(R \mid B) = \frac{\text{Number of red balls}}{\text{Total remaining balls}} = \frac{4}{11} \approx 0.3636 \][/tex]
### Step 3: Calculate [tex]\( P(B \cap R) \)[/tex]
To find [tex]\( P(B \cap R) \)[/tex], the joint probability of both events occurring (choosing a black ball first and a red ball second):
- This is found by multiplying the probabilities from the previous steps, since the events are sequential and conditional.
[tex]\[ P(B \cap R) = P(B) \times P(R \mid B) = \left( \frac{2}{3} \right) \times \left( \frac{4}{11} \right) = \frac{8}{33} \approx 0.2424 \][/tex]
### Summary
Here are the probabilities:
[tex]\[ P(B) = \frac{2}{3} \approx 0.6667 \][/tex]
[tex]\[ P(R \mid B) = \frac{4}{11} \approx 0.3636 \][/tex]
[tex]\[ P(B \cap R) = \frac{8}{33} \approx 0.2424 \][/tex]
Thus, the filled-in probabilities are:
[tex]\[
\begin{array}{l}
P(B) = 0.6667 \\
P(R \mid B) = 0.3636 \\
P(B \cap R) = 0.2424 \\
\end{array}
\][/tex]
