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: Use the formula for independent - events

For independent events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\). Given \(P(A)=\frac{1}{8}\) and \(P(A\cap B)=\frac{1}{12}\), we can find \(P(B)\). Since \(P(A\cap B) = P(A)\times P(B)\), then \(P(B)=\frac{P(A\cap B)}{P(A)}\).
\[P(B)=\frac{\frac{1}{12}}{\frac{1}{8}}=\frac{1}{12}\times\frac{8}{1}=\frac{2}{3}\]

Step2: Calculate \(P(B\cap C)\)

Assuming \(B\) and \(C\) are independent events, the formula for the probability of the intersection of two independent events is \(P(B\cap C)=P(B)\times P(C)\). We know \(P(B)=\frac{2}{3}\) and \(P(C)=\frac{1}{4}\).
\[P(B\cap C)=\frac{2}{3}\times\frac{1}{4}=\frac{1}{6}\]

Answer:

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