<p>Three coins are tossed once. Let A denote the event 'three heads show,' B denote the event 'two heads and one tail show', C denote the event 'three tails show' and D denote the event 'a head shows on the first toss', which events are mutually exclusive?</p>
<p>(a) \(A \cap B\), \(A \cap D\)</p>
<p>(b) \(A \cap C\), \(B \cap D\)</p>
<p>(c) \(B \cap C\), \(C \cap D\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Two events are mutually exclusive if their intersection is empty. Check each pair by finding if there are any common outcomes.
<p><strong>Step 1:</strong> When three coins are tossed, there are $2^3 = 8$ possible outcomes.</p><p>$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$</p><p><strong>Step 2:</strong> Identify the events:</p><ul><li>$A$ = Three heads show = $\{HHH\}$</li><li>$B$ = Two heads and one tail show = $\{HHT, HTH, THH\}$</li><li>$C$ = Three tails show = $\{TTT\}$</li><li>$D$ = A head shows on the first toss = $\{HHH, HHT, HTH, HTT\}$</li></ul><p><strong>Step 3:</strong> Check which pairs are mutually exclusive (i.e., $A_i \cap A_j = \emptyset$ for $i \neq j$):</p><ul><li>$B \cap C = \{HHT, HTH, THH\} \cap \{TTT\} = \emptyset$ ✓ (mutually exclusive)</li><li>$C \cap D = \{TTT\} \cap \{HHH, HHT, HTH, HTT\} = \emptyset$ ✓ (mutually exclusive)</li></ul><p>∴ Answer is (c).</p>
Correct Answer: c