Bihar Board Class 12 (Arts) Probability — practice questions — page 2

1. Your syllabus: two events are MUTUALLY EXCLUSIVE (or DISJOINT) when they cannot occur SIMULTANEOUSLY. 2. Formally: $A \cap B = \emptyset$ — no outcome lies in both events. 3. Consequence: $P(A \cap B) = 0$ for mutually exclusive events, and the addition rule simplifies to $P(A \cup B) = P(A) + P(B)$. 4. Examples: rolling a die — 'showing 1' and 'showing 6' are mutually exclusive; 'odd' and 'even' are mutually exclusive AND exhaustive. 5. Options A, B, D describe different relationships (equal probabilities, exhaustive together, identical events) — not mutual exclusion. _Source: syllabus Mathematics, Ch 14, §14.1 (Mutually exclusive events), p. 2._

1. Sample space: $\{HH, HT, TH, TT\}$ — $|S| = 4$ equally likely outcomes. 2. Event 'at least one head': occurs when at least one of the two coins shows H. Enumerate: $\{HH, HT, TH\}$ — three outcomes. 3. $P(\text{at least 1 H}) = 3/4$. 4. Alternative via COMPLEMENT: $P(\text{at least 1 H}) = 1 - P(\text{no H}) = 1 - P(\text{TT}) = 1 - 1/4 = 3/4$. (Often faster.) 5. Option A is $P(\text{exactly two heads}) = 1/4$. Option B is $P(\text{exactly one head}) = 1/2$. Option D would mean certainty. _Source: syllabus Mathematics, Ch 14, §14.2 (Events and probability calculation), p. 4–5._

1. 'Pass' and 'fail' are COMPLEMENTARY events: they are mutually exclusive AND exhaustive. 2. By the complement rule: $P(\text{fail}) = 1 - P(\text{pass})$. 3. Plug in: $P(\text{fail}) = 1 - 0.6 = 0.4$. 4. Sum check: $P(\text{pass}) + P(\text{fail}) = 0.6 + 0.4 = 1.0$ ✓ (since they cover all possibilities). 5. Option A would repeat $P(\text{pass})$. Options C, D are arithmetic errors. _Source: syllabus Mathematics, Ch 14, §14.2.2 (Complement of an event), p. 8–9._