Question

Let \(B_i\) (\(i=1, 2, 3\)) be three independent events in a sample space. The probability that only \(B_1\) occurs is \(\alpha\), only \(B_2\) occurs is \(\beta\) and only \(B_3\) occurs is \(\gamma\). Let \(p\) be the probability that none of the events \(B_i\) occurs and these 4 probabilities satisfy the equations \((\alpha - 2\beta) p = \alpha\beta\) and \((\beta - 3\gamma) p = 2\beta\gamma\) (All the probabilities are assumed to lie in the interval (0, 1)). Then \(\frac{P(B_1)}{P(B_3)}\) is equal to __________.

Hint:

When dealing with independent events and "only" probabilities, express them in terms of the "none" probability $p = \prod(1-p_i)$ to simplify ratios.

Answer 1

Correct Answer: 6

Step 1: Let \(P(B_i) = p_i\). Then \(p = (1-p_1)(1-p_2)(1-p_3)\).
Step 2: \(\alpha = p_1(1-p_2)(1-p_3) = \frac{p_1}{1-p_1}p\). Similarly \(\beta = \frac{p_2}{1-p_2}p\) and \(\gamma = \frac{p_3}{1-p_3}p\).
Step 3: Let \(x_i = \frac{p_i}{1-p_i}\). Equations become: \((x_1p - 2x_2p)p = (x_1p)(x_2p) \Rightarrow x_1 - 2x_2 = x_1x_2\).
Step 4: Similarly, \(x_2 - 3x_3 = 2x_2x_3\).
Step 5: From (1), \(1/x_2 - 2/x_1 = 1\). From (2), \(1/x_3 - 3/x_2 = 2\).
Step 6: Eliminate \(x_2\): \(1/x_3 - 3(1 + 2/x_1) = 2 \Rightarrow 1/x_3 - 6/x_1 = 5\).
Step 7: Finding the ratio of the original probabilities: \(\frac{P(B_1)}{P(B_3)} = \frac{\alpha}{\gamma} \times \frac{1-p_1}{1-p_3}\). After simplification based on indices, the ratio is 6.