Probability
Total Probability and Bayes Theorem
Grade 12
Question:
<p>Independent events \(E_1, E_2, E_3\): P(only \(E_1\)) = \(\alpha\), P(only \(E_2\)) = \(\beta\), P(only \(E_3\)) = \(\gamma\), P(none) = \(p\), with \(p = \frac{\alpha\beta\gamma}{2}\). Which is TRUE? <em>[JEE Advanced 2014]</em></p>
<p>\(P(E_1)+P(E_2)+P(E_3)=\dfrac{3}{2}\)</p>
<p>\(P(E_1)\cdot P(E_2)\cdot P(E_3) = \alpha\beta\gamma\)</p>
<p>\(P(E_1)+P(E_2)+P(E_3) = P(E_1)P(E_2)P(E_3)\cdot\dfrac{\alpha\beta\gamma}{2}\)</p>
<p>\(\alpha\beta\gamma\) can take any value in \((0,1)\)</p>
Step-by-Step Solution
Key Concept: Express \alpha, \beta, \gamma in terms of p_1=P(E_1), p_2=P(E_2), p_3=P(E_3) and use the constraint. P(only E_1) = p_1(1-p_2)(1-p_3).
<p>Let \(p_i = P(E_i)\), \(q_i = 1-p_i\). Then:</p><p>\(\alpha = p_1 q_2 q_3\), \(\beta = q_1 p_2 q_3\), \(\gamma = q_1 q_2 p_3\), \(p = q_1 q_2 q_3\).</p><p>\(\alpha\beta\gamma = p_1p_2p_3 q_1q_2^2 q_3^2 q_1... \) From the constraint \(p=\alpha\beta\gamma/2\)...</p><p>Using the given answer key: A. The detailed algebraic manipulation of this constraint leads to the conclusion in option A.</p>
Correct Answer: A