Probability
Assertion-Reason Type Problems
Grade 12
Question:
<p><strong>Example 29 (Assertion-Reason):</strong></p><p><strong>Statement-1:</strong> A man P speaks truth with probability $p$ and another man Q speaks truth with probability $2p$. If P and Q contradict each other with probability $\frac{1}{2}$, then there are two values of $p$.</p><p><strong>Statement-2:</strong> A quadratic equation with real coefficients has two real roots.</p>
<p>(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1</p>
<p>(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1</p>
<p>(c) Statement-1 is true, Statement-2 is false</p>
<p>(d) Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: Find the probability condition for contradiction and solve the resulting quadratic equation to determine the number of solutions for $p$.
<p><strong>Step 1:</strong> Let $E_1$ be the event that P speaks the truth, then $P(E_1) = p$. Let $E_2$ be the event that Q speaks the truth, then $P(E_2) = 2p$.</p><p><strong>Step 2:</strong> If P and Q contradict each other with probability $\frac{1}{2}$:</p><p>$$P(E_1) \cdot P(\bar{E_2}) + P(\bar{E_1}) \cdot P(E_2) = \frac{1}{2}$$</p><p>$$p(1-2p) + (1-p)(2p) = \frac{1}{2}$$</p><p>$$p - 2p^2 + 2p - 2p^2 = \frac{1}{2}$$</p><p>$$3p - 4p^2 = \frac{1}{2}$$</p><p>$$8p^2 - 6p + 1 = 0$$</p><p><strong>Step 3:</strong> Using the quadratic formula: $\Delta = 36 - 32 = 4 > 0$, so there are two distinct real values of $p$.</p><p><strong>Statement-1 is TRUE.</strong></p><p><strong>Step 4:</strong> Statement-2 claims a quadratic equation with real coefficients always has two real roots. This is false—it can have complex roots if the discriminant is negative.</p><p><strong>Statement-2 is FALSE.</strong></p><p>∴ Answer is (c).</p>
Correct Answer: C