Question

Let there be three independent events \(E_1\), \(E_2\) and \(E_3\). The probability that only \(E_1\) occurs is \(\alpha\), only \(E_2\) occurs is \(\beta\) and only \(E_3\) occurs is \(\gamma\). Let 'p' denote the probability of none of events occurs that satisfies the equations \((\alpha - 2\beta)p = \alpha\beta\) and \((\beta - 3\gamma)p = 2\beta\gamma\). All the given probabilities are assumed to lie in the interval \((0, 1)\).
Then, \(\dfrac{\text{Probability of occurrence of } E_1}{\text{Probability of occurrence of } E_3}\) is equal to _____________.

Hint:

Let \(P(E_i) = p_i\). Then \(\alpha/p = p_1/(1-p_1)\), etc. Convert both equations into relations among \(\frac{1-p_i}{p_i}\).

Answer 1

Correct Answer: 6

Let \(p_i = P(E_i)\). Independence gives \(\alpha = p_1(1-p_2)(1-p_3)\), \(\beta = (1-p_1)p_2(1-p_3)\), \(\gamma = (1-p_1)(1-p_2)p_3\) and \(p = (1-p_1)(1-p_2)(1-p_3)\).
So \(\frac{\alpha}{p} = \frac{p_1}{1-p_1}\), \(\frac{\beta}{p} = \frac{p_2}{1-p_2}\), \(\frac{\gamma}{p} = \frac{p_3}{1-p_3}\).
Let \(u = \frac{1-p_1}{p_1}\), \(v = \frac{1-p_2}{p_2}\), \(w = \frac{1-p_3}{p_3}\), so \(\frac{p}{\alpha} = u\), \(\frac{p}{\beta} = v\), \(\frac{p}{\gamma} = w\).
Equation 1: divide \((\alpha - 2\beta)p = \alpha\beta\) by \(\alpha\beta\): \(\frac{p}{\beta} - \frac{2p}{\alpha} = 1\Rightarrow v - 2u = 1\).
Equation 2: divide \((\beta - 3\gamma)p = 2\beta\gamma\) by \(\beta\gamma\): \(\frac{p}{\gamma} - \frac{3p}{\beta} = 2\Rightarrow w - 3v = 2\).
So \(v = 1 + 2u\), \(w = 2 + 3v = 5 + 6u\).
\(\frac{P(E_1)}{P(E_3)} = \frac{p_1}{p_3} = \frac{1/(1+u)}{1/(1+w)} = \frac{1+w}{1+u} = \frac{6 + 6u}{1 + u} = 6\).
Answer: 6