Question

Question: Suppose that \({E_1}\;,\;{E_2}\;, \ldots ,\;{E_n}\) are \(n\) events with \(p\left( {{E_i}} \right) > 0\) for \(i = 1\;,\;2\;, \ldots ,\;n\). Show that

\(P\left( {{E_1} \cap {E_2} \cap \cdots \cap {E_n}} \right) = P\left( {{E_1}} \right)P\left( {{E_2}\mid {E_1}} \right)P\left( {{E_3}\mid {E_1} \cap {E_2}} \right) \cdots P\left( {{E_n}\mid {E_1} \cap {E_2} \cap \cdots \cap {E_{n - 1}}} \right)\)

Short answer

Answer

The equation \(P\left( {{E_1} \cap {E_2} \cap \ldots .. \cap {E_n}} \right) = P\left( {{E_1}} \right)P\left( {{E_2}\mid {E_1}} \right)P\left( {{E_3}\mid {E_1} \cap {E_2}} \right) \ldots ..P\left( {{E_n}\mid {E_1} \cap {E_2} \cap \ldots ..{E_{n - 1}}} \right)\) has been proved.

Step-by-step solution

Given data 

The \(n\) events with the probability \(P\left( {{E_i}} \right) > 0\) for \(i = 1\;,\;2\;, \ldots \) are \({E_1}\;,\;{E_2}\;,\;{E_3}\;, \ldots \ldots \ldots {E_n}\).

Concept of Conditional probability

The probability of occurrence of any event \({\bf{A}}\) when another event \(B\) in relation to \({\bf{A}}\) has already occurred is known as conditional probability. It is depicted by \({\rm{P}}({\rm{A}}\mid {\rm{B}})\).

Calculation for the probability of three events

Consider two events \({E_1},{E_2}\).

With the use of conditional probability, we have \(P\left( {{E_1} \cap {E_2}} \right) = P\left( {{E_1}} \right)P\left( {{E_2}\mid {E_1}} \right)\).

Consider three events \({E_1}\;,\;{E_2}\;,\;{E_3}\) with the use of associative property:

\(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = P\left\{ {{E_1} \cap \left( {{E_2} \cap {E_3}} \right)} \right\}\)

If \({E_2} \cap {E_3} = {E_t}\), then it implies,

\(\begin{aligned}{}P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) &= P\left\{ {{E_1} \cap {E_t}} \right\}\\P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) &= P\left( {{E_1}} \right)P\left( {{E_t}\mid {E_1}} \right)\end{aligned}\)

\(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = P\left( {{E_1}} \right)P\left( {\left( {{E_2} \cap {E_3}} \right)\mid {E_1}} \right)\) …… (1)

\(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = P\left( {{E_2} \cap {E_3}} \right)P\left( {{E_1}/{E_2} \cap {E_3}} \right)\) …… (2)

From equation (1) and (2):

\(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = P\left( {{E_3}} \right) \cdot P\left( {{E_2}/{E_3}} \right) \cdot P\left( {{E_1}/{E_1} \cap {E_2}} \right)\)

From equations (1) and (2):

\(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = P\left( {{E_1}} \right) \cdot P\left( {{E_2}/{E_1}} \right) \cdot P\left( {\left( {{E_3}} \right)\mid {E_1} \cap {E_2}} \right)\)

Calculation for the probability of four events

Consider the four events \({E_1}\;,\;{E_2}\;,\;{E_3}\;,\;{E_4}\).

\(P\left( {{E_1} \cap {E_2} \cap {E_3} \cap {E_4}} \right) = P\left( {{E_1}} \right) \cdot P\left( {{E_2}/{E_1}} \right) \cdot P\left( {\left( {{E_3}} \right)\mid {E_1} \cap {E_2}} \right) \cdot P\left( {\left( {{E_4}/{E_1} \cap {E_2} \cap {E_3}} \right)} \right.\)

Similarly, for \(n\) number of events:

\(P\left( {{E_1} \cap {E_2} \cap \ldots .. \cap {E_n}} \right) = P\left( {{E_1}} \right)P\left( {{E_2}\mid {E_1}} \right)P\left( {{E_3}\mid {E_1} \cap {E_2}} \right) \ldots ..P\left( {{E_n}\mid {E_1} \cap {E_2} \cap \ldots ..{E_{n - 1}}} \right)\)

Hence, proved.