Question

To Determine a formula for the probability of \({E_1} \cup {E_2} \cup {E_3}\).

Short answer

The probability formula \(p\left( {{E_1} \cup {E_2} \cup {E_3}} \right) = \sum\limits_{i = 1}^3 p \left( {{E_i}} \right) - \sum\limits_{1 \le i < j \le 3} p \left( {{E_i} \cap {E_j}} \right) + p\left( {{E_1} \cap {E_2} \cap {E_3}} \right)\)

Step-by-step solution

 Given

The condition of probability \({E_1} \cup {E_2} \cup {E_3}\).

The Concept of Principle of inclusion-exclusion

Principle of inclusion-exclusion:

\(\left| {{A_1} \cup {A_2} \cup \ldots \cup {A_n}} \right| = \sum\limits_{1 \le i \le n} {\left| {{A_i}} \right|} - \sum\limits_{1 \le i < j \le n} {\left| {{A_i} \cap {A_j}} \right|} + \sum\limits_{1 \le i < j < k \le n} {\left| {{A_i} \cap {A_j} \cap {A_k}} \right|} \quad - \ldots . + \left| {{A_1} \cap {A_2} \cap \ldots . \cap {A_n}} \right|\)

Determine the probability Formula

Principle of inclusion-exclusion:

\(\left| {{A_1} \cup {A_2} \cup \ldots \cup {A_n}} \right| = \sum\limits_{1 \le i \le n} {\left| {{A_i}} \right|} - \sum\limits_{1 \le i < j \le n} {\left| {{A_i} \cap {A_j}} \right|} + \sum\limits_{1 \le i < j < k \le n} {\left| {{A_i} \cap {A_j} \cap {A_k}} \right|} \quad - \ldots . + \left| {{A_1} \cap {A_2} \cap \ldots . \cap {A_n}} \right|\)

The probability of any event \(E \subseteq S\) is \(p\left( {{E_i}} \right) = \frac{{|E|}}{{|S|}}\).

Thus, using the inclusion-exclusion principle:

\(\begin{aligned}{c}p\left( {{E_1} \cup {E_2} \cup {E_3}} \right) &= \frac{{\left| {{E_1} \cup {E_2} \cup {E_3}} \right|}}{{|S|}}\\ &= \frac{{\left| {{E_1}} \right| + \left| {{E_2}} \right| + \left| {{E_3}} \right| - \left| {{E_1} \cap {E_2}} \right| - \left| {{E_1} \cap {E_3}} \right| - \left| {{E_2} \cap {E_3}} \right| + \left| {{E_1} \cap {E_2} \cap {E_3}} \right|}}{{|S|}}\\p\left( {{E_1} \cup {E_2} \cup {E_3}} \right) &= \sum\limits_{i = 1}^3 p \left( {{E_i}} \right) - \sum\limits_{1 \le i < j \le 3} p \left( {{E_i} \cap {E_j}} \right) + p\left( {{E_1} \cap {E_2} \cap {E_3}} \right)\end{aligned}\)