Question

Question:Suppose that at least one of the events\({E_j},j = 1,2, \ldots ,m\), is guaranteed to occur and no more than two can occur. Show that if\(p\left( {{E_j}} \right) = q\)for\(j = 1,2, \ldots ,m\)and\(p\left( {{E_j} \cap {E_k}} \right) = r\)for\(1 \le j < k \le m\), then\(q \ge 1/m\)and\(r \le 2/m\).

Short answer

Answer

Let\(X = \sum\limits_{j = 1}^m {{X_j}} \)be a random variable where\(X\_\left\{ j \right\} = 1,\)if\({E_j}\)occurs and 0 otherwise\(\forall j = 1,2, \ldots ,m.\) Obviously\(E\left( {{X_j}} \right) = P\left( {{E_j}} \right) = q\forall j\). Thus

\(E(X) = \sum\limits_{j = 1}^m E \left( {{X_j}} \right) = qm\).

Step-by-step solution

Formula for sum of Probabilities 

The probability of the event\(E\)is the sum of the probabilities of the outcomes in\(E\). That is,\(p(E) = \sum\limits_{s \in E} p (s).\)

(Note that when \(E\) is an infinite set, \(\sum\limits_{s \in E} p (s)\) is a convergent infinite series.)

Use Probability sum

Let\(X = \sum\limits_{j = 1}^m {{X_j}} \)be a random variable where\({X_j} = 1\), if\({E_j}\)occurs and\(0\)otherwise\(\forall j = 1,2, \ldots ,m\). Obviously\(E\left( {{X_j}} \right) = P\left( {{E_j}} \right) = q\forall j\).Thus

\(E(X) = \sum\limits_{j = 1}^m E \left( {{X_j}} \right) = qm\)

Since one of the events\({E_j}\) is certain to occur, so\(E(X) \ge 1 \Rightarrow q \ge \frac{1}{m}.\)

Again we observe that as one of the events is certain to occur and no more than two can occur so

\(P\left( {\bigcup\limits_{i = 1}^m {{E_j}} } \right) = \sum\limits_{i = 1}^m P \left( {{E_j}} \right) - \sum\limits_{i < j} P \left( {{E_i} \cap {E_j}} \right) = q \cdot m - \left( {\begin{aligned}{*{20}{c}}m\\2\end{aligned}} \right) \cdot r = 1\)

It follows using\(q \le 1\) that \(r = \frac{{gm - 1}}{{\left( {\begin{aligned}{*{20}{c}}m\\2\end{aligned}} \right)}} = \frac{2}{m} \cdot \frac{{gm - 1}}{{m - 1}} \le \frac{2}{m}.\)