NCERT Solutions for Class 12 Maths Chapter 13 – Probability Exercise 13.2

Flashcard for Question 1

Quick Tip: Use independence: P(A ∩ B) = P(A) × P(B).

Common Mistake: Adding probabilities or using the union formula instead of multiplying; forgetting this works only if A and B are independent.

Flashcard for Question 2

Quick Tip: Without replacement → multiply sequential probabilities: (26/52) × (25/51) = 25/102.

Common Mistake: Treating draws as with replacement (using 26/52 twice), or forgetting to reduce the denominator on the second draw

Flashcard for Question 3

Quick Tip: For “all good” without replacement, multiply sequential probabilities: (12/15) × (11/14) × (10/13).

Common Mistake: Treating draws as with replacement (using 12/15 three times) or accidentally including outcomes with a bad orange.

Flashcard for Question 4

Quick Tip: Check independence using P(A ∩ B) = P(A) × P(B). Calculate each probability separately and compare.

Common Mistake: Assuming independence without verification or mixing up with mutually exclusive events (they’re different concepts).

Flashcard for Question 5

Quick Tip: Compute P(A), P(B), and P(A ∩ B). Then check if P(A ∩ B) = P(A) × P(B).

Common Mistake: Confusing colour grouping with even/odd grouping, or forgetting to divide by 6 (total outcomes).

Exam Insight: Always write probabilities as fractions over total outcomes before comparing—this shows independence clearly and avoids silly mistakes.

Flashcard for Question 6

Quick Tip: For independence, check if P(E ∩ F) = P(E) × P(F).

Common Mistake: Just comparing P(E) + P(F) with 1 or assuming independence without checking the product rule.

Exam Insight: Write down P(E), P(F), and their product first—if it matches the given P(E ∩ F), then independent; otherwise, not. This direct method saves time in exams.

Flashcard for Question 9

Quick Tip: Use De Morgan’s law: P(not A and not B) = 1 – P(A ∪ B). Then apply P(A ∪ B) = P(A) + P(B) – P(A ∩ B).

Common Mistake: Forgetting to subtract P(A ∩ B) when finding P(A ∪ B), leading to double counting.

Exam Insight: Always reduce the problem to union/intersection formulas first – it’s faster and avoids mistakes in “not A and not B” type questions.

Flashcard for Question 10

Quick Tip: Convert “not A or not B” into 1 – P(A ∩ B). Then compare P(A ∩ B) with P(A) × P(B) to test independence.

Common Mistake: Mixing up “not A or not B” with “not (A ∩ B)” – careless handling of complements often leads to wrong probability.

Flashcard for Question 12

Quick Tip: Use complement: P(at least once odd) = 1 – P(no odd in 3 tosses).

Common Mistake: Trying to count all favourable cases directly instead of using complement, which is much simpler.

Exam Insight: In “at least once” problems, always think of the complement—it saves time and avoids messy casework.

Flashcard for Question 13

Quick Tip: Since it’s with replacement, probabilities stay the same for both draws. Use multiplication for sequential events.

Common Mistake: Forgetting replacement rule—many students reduce the denominator on the second draw, which is only correct for without replacement.

Flashcard for Question 14

Quick Tip: Use complement for part (i): P(solved) = 1 – P(both fail). For part (ii), add probabilities of “A solves, B fails” and “B solves, A fails.”

Common Mistake: Forgetting independence when multiplying probabilities of success/failure, or double counting the case when both solve.

Flashcard for Question 16

Quick Tip: Use Venn diagram logic. Apply inclusion–exclusion: P(H ∪ E) = P(H) + P(E) – P(H ∩ E). Then use conditional probability for (b) and (c).

Common Mistake: Forgetting to subtract the intersection when finding union, or mixing up conditional probability formula P(A|B) = P(A ∩ B)/P(B).

Exam Insight: Always calculate the “neither” case first (complement of union). For conditional parts, directly plug into the formula – this is a common exam scoring area where clarity matters.