Discussion
There are four cities, namely, \(C_1\), \(C_2\), \(C_3\) and \(C_4\). The cities are directly connected by four roads as shown in the picture given below, that is, \(C_1\) is connected with \(C_2\), \(C_2\) is connected with \(C_3\), \(C_3\) is connected with \(C_4\), and \(C_4\) is connected with \(C_1\). The probability of any road getting independently blocked is \(\frac{1}{3}\). Let \(E_1\) be the event of travelling from \(C_1\) to \(C_3\) via \(C_2\) and \(E_2\) be the event of travelling from \(C_1\) to \(C_3\) via \(C_4\). Then, which of the following statements is correct ?
(Figure: the four cities at the corners of a square, joined by the roads \(C_1C_2\), \(C_2C_3\), \(C_3C_4\) and \(C_4C_1\).)
Answer: A
The correct statement is \(P(E_1 \cup E_2) = \frac{56}{81}\).
Each road is open with probability \(2/3\). Travelling via \(C_2\) needs both roads \(C_1C_2\) and \(C_2C_3\) open, so \(P(E_1) = (2/3)^{2} = 4/9\), and likewise \(P(E_2) = 4/9\). The two routes use different roads, so the events are independent:
\[ \begin{aligned} P(E_1 \cup E_2) &= \tfrac{4}{9} + \tfrac{4}{9} - \tfrac{4}{9} \cdot \tfrac{4}{9} \\ &= \tfrac{72}{81} - \tfrac{16}{81} = \tfrac{56}{81} \end{aligned} \]
\(\frac{8}{9}\) adds the two probabilities without removing the overlap. Independence means \(P(E_1|E_2) = P(E_1)\), so option C is false, and \(P(E_1 \cap E_2) = 16/81\), not 0.
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