Let us consider an example, where we flip 3 different fair coins and the sample space is:

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Then probability that the first coin comes up heads = ({HHH, HHT, HTH, HTT }) = \(\cfrac{4}{8} = \cfrac{1}{2}\)

Now suppose we have an additional information that at least 2 of the coins are showing heads, this informant has changed our available information, now the sample space will be reduced to ({HHH, HHT, HTH, THH}). Out of these 4 equally likely outcomes, there are 3 cases when first coin is showing head.  Hence the probability that the first coin is heads, given that at least two of the three coins are heads is \(\frac{3}{4}\). This is an example of conditional probability. We use a mathematical notation 

\[P({\rm{First}}\;{\rm{coin}}\;{\rm{head}}\;{\rm{|}}\;{\rm{at}}\;{\rm{least}}\;{\rm{2}}\;{\rm{heads}}) = \frac{3}{4}\]

Given two events \(A\) and \(B\), with \(P(B) > 0\), the conditional probability of \(A\) given that \(B\) has happened is equal to:

 \[P(A|B) = \frac{{P(A \cap B)}}{{P(B)}}\]

Three dice are thrown, find the probability that the sum of the outcomes is 16, given that the outcome of one of the dice is 5.