calculating the probability of independent events\nconsider a situation in which $p(a)=\frac{1}{8}$…

calculating the probability of independent events\nconsider 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 $p(b \text{ and } c)?$

calculating the probability of independent events\nconsider 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 $p(b \text{ and } c)?$

Answer

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})