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For two events $A$ and $B$, the conditional probability of event $A$, is the probability of event $A$ given the occurence (or non-occurence) of event $B$.
The conditional probability of $A$ given $B$, written as $\mathbb{P}(A\mid B)$, is defined by
$$\mathbb{P}(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}$$If events $A$ and $B$ are independent, $\mathbb{P}(A\cap B)=\mathbb{P}(A)\times \mathbb{P}(B)$ . Hence
$$\mathbb{P}(A\mid B)=\mathbb{P}(A)$$ $$\mathbb{P}(B\mid A)=\mathbb{P}(B)$$In other words, the probability of $A$ given $B$ is just the probability of $A$ (regardless of $B$ — whether $B$ has occured or not doesn't affect the probability of $A$, because they are independent events).
Examples:
In a certain region, on any day of the year, the probability that it's cloudy is 0.4. It's also known that there's a 0.3 probability that any day is a cloudy and rainy day. Given that today is cloudy, what is the probability that it will rain?
Let $C$ be the event that it's cloudy and $R$ be the event that it rains.
$$\begin{array}{rl}\mathbb{P}(C)& =0.4\\ \mathbb{P}(R\cap C)& =0.3\\ \mathbb{P}(R\mid C)& =\frac{\mathbb{P}(R\cap C)}{\mathbb{P}(C)}\\ & =\frac{0.3}{0.4}\\ & =0.75\end{array}$$Given today is cloudy, there's a 75% chance it will rain today.
In a city, the ratio between men and women is 6:4. Thirty percent of the men are vegetarian. What percentage of the city residents are vegetarian men?
Let $M$ be the probability of any random resident being a man and $V$ be the probability that any random resident being a vegetarian.
$$\begin{array}{rl}\mathbb{P}(M)& =0.6\\ \mathbb{P}(V\mid M)& =0.3\\ \mathbb{P}(V\mid M)& =\frac{\mathbb{P}(V\cap M)}{\mathbb{P}(M)}\\ 0.3& =\frac{\mathbb{P}(V\cap M)}{0.6}\\ \mathbb{P}(V\cap M)& =0.3\times 0.6\\ \mathbb{P}(V\cap M)& =0.18\end{array}$$Eighteen percent of the city residents are vegetarian men.
See also: numbers, Permutations and Combinations