NCERT Solutions for Class 12 Maths Chapter 13 – Probability Exercise 13.1

Flashcard for Question 1

Quick Tip: Use conditional probability: P(E|F) = P(E ∩ F) / P(F) and P(F|E) = P(E ∩ F) / P(E).

Common Mistake: Swapping numerator/denominator or using P(E ∪ F) instead of P(E ∩ F) in the formula.

Exam Insight: Check that the conditioning event has nonzero probability first; then simplify fractions exactly to avoid rounding errors.

Flashcard for Question 2

Quick Tip: Use P(A|B) = P(A ∩ B) / P(B) directly—plug in and simplify.

Common Mistake: Dividing by P(A) instead of P(B), or forgetting that the denominator must be > 0.

Exam Insight: Write the formula first, then put in the numbers. This keeps you from switching numbers by accident and gets you method marks even if you make a mistake in math.

Flashcard for Question 3

Quick Tip: Start with P(A ∩ B) = P(B|A) × P(A), then get P(A|B) = P(A ∩ B) / P(B), and finally use P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Common Mistake: Swapping P(B|A) with P(A|B), or forgetting to subtract the intersection when computing the union.

Exam Insight: Do the steps in that order to avoid backtracking, and sanity-check that 0 ≤ each probability ≤ 1 and P(A ∪ B) ≤ 1.

Flashcard for Question 4

Quick Tip: Use P(A|B) = P(A ∩ B)/P(B) to find P(A ∩ B), then apply P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Common Mistake: Forgetting that “2P(A) = P(B)” gives a relation between P(A) and P(B) and must be used before substitution.

Flashcard for Question 5

Quick Tip: Use formula P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to get intersection first, then apply conditional probability definitions.

Common Mistake: Mixing up conditional probabilities (P(A|B) vs P(B|A)) or forgetting to divide by the correct denominator.

Flashcard for Question 6

Quick Tip: Write the full sample space (8 outcomes) first. Then list outcomes for E and F carefully, and use them to check intersections/unions if needed.

Common Mistake: Missing cases in “at least” or “at most” type events—students often drop boundary cases like exactly 2 heads or exactly 2 tails.

Exam Insight: Always count systematically (e.g., by number of heads/tails) rather than writing outcomes randomly – this minimizes double-counting and saves time in exams.

Flashcard for Question 7

Quick Tip: Write the sample space {HH, HT, TH, TT}. Then pick outcomes that match E and F, and check their relation.

Common Mistake: Misinterpreting “one coin shows head” as “exactly one head” instead of “at least one head.” Similarly, mixing up “no tail” with “exactly one tail.”

Exam Insight: Carefully translate wording into outcomes before solving – exam questions often test precision in interpreting phrases like “one,” “at least one,” or “no.”

Flashcard for Question 8

Quick Tip: Treat each toss as independent – just multiply the probabilities of required outcomes for E and F.

Common Mistake: Forgetting independence and trying to adjust denominators across tosses, or ignoring the word “respectively” (order matters).

Flashcard for Question 9

Quick Tip: List all 3! = 6 arrangements and then count which ones satisfy E, F, or both.

Common Mistake: Forgetting that “son on one end” means either left or right, not just one side.

Flashcard for Question 10

Quick Tip: Fix the die value given in the condition first, then only vary the other die and count favourable outcomes over total possible in that reduced sample space.

Common Mistake: Forgetting to reduce the sample space after the condition (still using 36 outcomes instead of the restricted ones).

Flashcard for Question 11

Quick Tip: Use conditional probability formula P(A|B) = P(A ∩ B)/P(B). First write E, F, G sets, then find intersections/unions, and divide by size of the conditioning event.

Common Mistake: Forgetting to restrict the sample space to the conditioning event (using 6 as denominator instead of |B|).

Flashcard for Question 12

Quick Tip: Write the sample space {BB, BG, GB, GG}. Then apply the condition (youngest girl / at least one girl), restrict the sample space, and count favourable outcomes.

Flashcard for Question 13

Quick Tip: Use conditional probability: P(Easy | MCQ) = (Easy ∩ MCQ) / (MCQ). Count only multiple-choice questions in the denominator.

Flashcard for Question 14

Quick Tip: First restrict the sample space to outcomes with different numbers (36 − 6 = 30). Then count favourable ones for sum = 4.

Common Mistake: Forgetting to exclude doubles like (2,2), which changes both numerator and denominator.

Flashcard for Question 15

Quick Tip: Break into cases: first die is 3 (→ roll again) or not 3 (→ coin toss). Then apply the condition “at least one die shows 3” to restrict the sample space.

Common Mistake: Forgetting to condition properly—many students include coin outcomes even when both dice show 3 (no coin is tossed).

Exam Insight: In conditional probability word problems, always (1) define the reduced sample space clearly, (2) identify when the coin is actually tossed, and (3) then count favourable over total. This structured approach avoids confusion and earns method marks.