Certainly! Let's solve this step by step:
1. Given Information:
- [tex]\( P(A) = \frac{1}{8} \)[/tex]
- [tex]\( P(C) = \frac{1}{4} \)[/tex]
- [tex]\( P(A \text{ and } B) = \frac{1}{12} \)[/tex]
2. Calculate [tex]\( P(B) \)[/tex]:
- We know that [tex]\( P(A \text{ and } B) = P(A) \times P(B|A) \)[/tex]
- Assuming that events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent, then [tex]\( P(B|A) = P(B) \)[/tex]
- Thus, we can write: [tex]\( P(A \text{ and } B) = P(A) \times P(B) \)[/tex]
Given [tex]\( P(A \text{ and } B) = \frac{1}{12} \)[/tex] and [tex]\( P(A) = \frac{1}{8} \)[/tex], substitute these values into the equation:
[tex]\[
\frac{1}{12} = \frac{1}{8} \times P(B)
\][/tex]
Solve for [tex]\( P(B) \)[/tex]:
[tex]\[
P(B) = \frac{1}{12} \div \frac{1}{8} = \frac{1}{12} \times \frac{8}{1} = \frac{8}{12} = \frac{2}{3}
\][/tex]
3. Calculate [tex]\( P(B \text{ and } C) \)[/tex]:
- Now we need to find the probability of [tex]\( P(B \text{ and } C) \)[/tex]
- Assuming that events [tex]\( B \)[/tex] and [tex]\( C \)[/tex] are independent, we can use the formula: [tex]\( P(B \text{ and } C) = P(B) \times P(C) \)[/tex]
Given [tex]\( P(B) = \frac{2}{3} \)[/tex] and [tex]\( P(C) = \frac{1}{4} \)[/tex], substitute these values into the equation:
[tex]\[
P(B \text{ and } C) = \frac{2}{3} \times \frac{1}{4} = \frac{2 \times 1}{3 \times 4} = \frac{2}{12} = \frac{1}{6}
\][/tex]
Therefore, [tex]\( P(B \text{ and } C) = \frac{1}{6} \)[/tex].
