Certainly! Let's tackle each part of the problem step-by-step.
First, let's define the events:
- Let event [tex]\( B \)[/tex] be choosing a black ball first.
- Let event [tex]\( R \)[/tex] be choosing a red ball second, given that the first ball chosen was black.
1. Probability of choosing a black ball first, [tex]\( P(B) \)[/tex]:
The probability [tex]\( P(B) \)[/tex] represents the likelihood of selecting a black ball on the first draw. Given our information, we determine that:
[tex]\[
P(B) = \frac{7}{12} \approx 0.5833
\][/tex]
2. Conditional probability of choosing a red ball second, given that the first ball was black, [tex]\( P(R \mid B) \)[/tex]:
Given that the first ball chosen was black, the probability [tex]\( P(R \mid B) \)[/tex] indicates the likelihood of picking a red ball on the second draw. After drawing the first black ball, there are fewer total balls and black balls available. Thus:
[tex]\[
P(R \mid B) = \frac{5}{11} \approx 0.4545
\][/tex]
3. Joint probability of both events happening, [tex]\( P(B \cap R) \)[/tex]:
The joint probability [tex]\( P(B \cap R) \)[/tex] represents the likelihood of choosing a black ball first and then a red ball second. We calculate this by multiplying the individual probabilities:
[tex]\[
P(B \cap R) = P(B) \times P(R \mid B)
\][/tex]
Substituting the values we determined:
[tex]\[
P(B \cap R) = 0.5833 \times 0.4545 \approx 0.2652
\][/tex]
4. Converting the joint probability to a percentage:
To express the probability that the first ball chosen is black and the second ball chosen is red as a percentage, we multiply by 100:
[tex]\[
\text{Percentage Probability} = P(B \cap R) \times 100 \approx 26.52\%
\][/tex]
Summarizing all these probabilities:
[tex]\[
\begin{array}{l}
P(B) = 0.5833 \\
P(R \mid B) = 0.4545 \\
P(B \cap R) = 0.2652
\end{array}
\][/tex]
Therefore, the probability that the first ball chosen is black and the second ball chosen is red is about [tex]\( 26.52\% \)[/tex].
