Question

calculating the probability of independent events
consider a situation in which ( p(a) = \frac{1}{8} ), ( p(c) = \frac{1}{4} ), and ( p(a \text{ and } b) = \frac{1}{12} ). what is ( p(b \text{ and } c) )?

Explanation:

Step1: Recall the formula for independent events

For independent events \( A \) and \( B \), the probability of both occurring is \( P(A \text{ and } B)=P(A)\times P(B) \). We can use this to find \( P(B) \) first.
Given \( P(A)=\frac{1}{8} \) and \( P(A \text{ and } B)=\frac{1}{12} \), from \( P(A \text{ and } B)=P(A)\times P(B) \), we can solve for \( P(B) \):
\( P(B)=\frac{P(A \text{ and } B)}{P(A)}=\frac{\frac{1}{12}}{\frac{1}{8}}=\frac{1}{12}\times\frac{8}{1}=\frac{2}{3} \)

Step2: Calculate \( P(B \text{ and } C) \)

Now, for independent events \( B \) and \( C \), the formula \( P(B \text{ and } C)=P(B)\times P(C) \) applies. We know \( P(B)=\frac{2}{3} \) and \( P(C)=\frac{1}{4} \), so:
\( P(B \text{ and } C)=\frac{2}{3}\times\frac{1}{4}=\frac{2\times1}{3\times4}=\frac{1}{6} \)

Answer:

\(\frac{1}{6}\)