Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 12 Maths NCERT Solutions for the Chapter 13 Miscellaneous Exercise — all 13 questions solved, pulling together conditional probability, independence, and Bayes' theorem from across the whole chapter.
Questions 1–3 revisit conditional probability with a subset/null-set special case, a two-children puzzle, and a base-rate problem comparing grey hair in men versus women. Questions 4 and 5 step outside conditional probability into Bernoulli-trial and equally-likely calendar reasoning — the probability that at most 6 of 10 randomly chosen people are right-handed, and the chance a leap year contains 53 Tuesdays. Questions 6, 7 and 10 are full Bayes' theorem problems dressed in different scenarios — four boxes of coloured marbles, a patient's heart-attack risk under two different treatments, and a ball transferred between two bags before a draw. Question 8 asks for the probability that a randomly filled 2×2 determinant turns out positive, and Question 9 works with the failure probabilities of two parts in an electronic assembly. The exercise — and the chapter — closes with three MCQs testing the formal relationships between P(A|B), P(B|A), P(A) and P(B).
(i) If A\subset B, every outcome of A is also in B, so A\cap B=A. Then P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A)}{P(A)}=1.
(ii) If A\cap B=\varnothing, then P(A\cap B)=0, so P(B|A)=\dfrac{0}{P(A)}=0.
Writing each outcome as (elder, younger), the sample space is S=\{bb,bg,gb,gg\}, each with probability \dfrac14.
(i) "At least one male" is F=\{bb,bg,gb\}, so P(F)=\dfrac34. "Both male" is E=\{bb\}, so E\cap F=\{bb\} and P(E\cap F)=\dfrac14. Thus P(E|F)=\dfrac{1/4}{3/4}=\dfrac13.
(ii) "Elder child is female" is F=\{gb,gg\}, so P(F)=\dfrac12. "Both female" is E=\{gg\}, so E\cap F=\{gg\} and P(E\cap F)=\dfrac14. Thus P(E|F)=\dfrac{1/4}{1/2}=\dfrac12.
Let M: person is male, W: person is female, with P(M)=P(W)=\dfrac12. Let G: person has grey hair, with P(G|M)=0.05 and P(G|W)=0.0025.
By Bayes' theorem, P(M|G)=\dfrac{\frac12\times0.05}{\frac12\times0.05+\frac12\times0.0025}=\dfrac{0.025}{0.025+0.00125}=\dfrac{0.025}{0.02625}=\dfrac{20}{21}.
Let X be the number of right-handed people among 10 independently chosen people. Since each person is right-handed with a fixed probability p=0.9 independently of the others, X follows a binomial distribution: P(X=k)=\binom{10}{k}(0.9)^k(0.1)^{10-k}.
It's quicker to work through the complement: P(X\le6)=1-P(X\ge7)=1-\left[P(7)+P(8)+P(9)+P(10)\right].
P(X\ge7)=\binom{10}{7}(0.9)^7(0.1)^3+\binom{10}{8}(0.9)^8(0.1)^2+\binom{10}{9}(0.9)^9(0.1)+\binom{10}{10}(0.9)^{10}, which evaluates to approximately 0.0574+0.1937+0.3874+0.3487\approx0.9872.
So P(X\le6)\approx1-0.9872=0.0128.
A leap year has 366 days, which is exactly 52 weeks plus 2 extra days. The 52 complete weeks already contain exactly 52 of every weekday, so 53 Tuesdays occur exactly when Tuesday is one of these 2 leftover days.
The 2 leftover days form one of 7 equally likely consecutive pairs: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). Tuesday appears in exactly 2 of these 7 pairs — (Mon, Tue) and (Tue, Wed).
| Box | Red | White | Black | Total |
|---|---|---|---|---|
| A | 1 | 6 | 3 | 10 |
| B | 6 | 2 | 2 | 10 |
| C | 8 | 1 | 1 | 10 |
| D | 0 | 6 | 4 | 10 |
Each box is chosen with prior probability \dfrac14. Let R denote "the marble drawn is red." Then P(R|A)=\dfrac{1}{10}, P(R|B)=\dfrac{6}{10}, P(R|C)=\dfrac{8}{10}, P(R|D)=0.
P(R)=\dfrac14\left(\dfrac{1}{10}+\dfrac{6}{10}+\dfrac{8}{10}+0\right)=\dfrac14\times\dfrac{15}{10}=\dfrac{3}{8}.
By Bayes' theorem: P(A|R)=\dfrac{\frac14\times\frac{1}{10}}{\frac38}=\dfrac{1/40}{3/8}=\dfrac{1}{15}; P(B|R)=\dfrac{\frac14\times\frac{6}{10}}{\frac38}=\dfrac{6/40}{3/8}=\dfrac{2}{5}; P(C|R)=\dfrac{\frac14\times\frac{8}{10}}{\frac38}=\dfrac{8/40}{3/8}=\dfrac{8}{15}.
Let E_1: patient follows meditation and yoga, E_2: patient follows the drug, with P(E_1)=P(E_2)=0.5. Let A: patient suffers a heart attack.
Meditation and yoga reduces the 40% base risk by 30%, leaving 70% of it: P(A|E_1)=0.40\times0.70=0.28. The drug reduces the risk by 25%, leaving 75% of it: P(A|E_2)=0.40\times0.75=0.30.
By Bayes' theorem, P(E_1|A)=\dfrac{0.5\times0.28}{0.5\times0.28+0.5\times0.30}=\dfrac{0.14}{0.14+0.15}=\dfrac{0.14}{0.29}=\dfrac{14}{29}.
Let the determinant be \begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc, with each of a, b, c, d independently equal to 0 or 1. There are 2^4=16 equally likely combinations. Since each entry is 0 or 1, ad-bc can only be -1, 0 or 1, so "positive" means ad-bc=1, which requires ad=1 (so a=d=1) and bc=0.
With a=d=1 fixed, b and c range over 4 equally likely pairs: (0,0), (0,1), (1,0), (1,1). Of these, bc=0 for 3 pairs and bc=1 (giving determinant 0, not positive) for the remaining pair.
So exactly 3 of the 16 total combinations give a positive determinant.
"B fails alone" means B fails while A doesn't, i.e. P(B\cap A')=0.15. Since "B fails" overall is the union of "B fails alone" and "A and B both fail," P(B)=P(B\cap A')+P(A\cap B)=0.15+0.15=0.30.
(i) P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.15}{0.30}=0.5.
(ii) "A fails alone" means P(A\cap B')=P(A)-P(A\cap B)=0.2-0.15=0.05.
Let E_1: the transferred ball is red, E_2: the transferred ball is black. From Bag I, P(E_1)=\dfrac37 and P(E_2)=\dfrac47.
If a red ball is transferred, Bag II now has 5 red and 5 black (10 total), so P(\text{red drawn}|E_1)=\dfrac{5}{10}=\dfrac12. If a black ball is transferred, Bag II now has 4 red and 6 black (10 total), so P(\text{red drawn}|E_2)=\dfrac{4}{10}=\dfrac25.
By Bayes' theorem, P(E_2|\text{red drawn})=\dfrac{\frac47\times\frac25}{\frac37\times\frac12+\frac47\times\frac25}=\dfrac{8/35}{3/14+8/35}. Using a common denominator of 70: \dfrac{16/70}{15/70+16/70}=\dfrac{16}{31}.
P(B|A)=1 means \dfrac{P(A\cap B)}{P(A)}=1, so P(A\cap B)=P(A). Since A\cap B\subset A always, and here it has the same probability as A itself, every outcome of A must (up to a null set) lie in B — that is, A\subset B.
P(A|B) \gt P(A) means \dfrac{P(A\cap B)}{P(B)} \gt P(A), so P(A\cap B) \gt P(A)\cdot P(B). Dividing both sides by P(A) gives \dfrac{P(A\cap B)}{P(A)} \gt P(B), i.e. P(B|A) \gt P(B) — the relationship is symmetric.
Rearranging P(A)+P(B)-P(A\cap B)=P(A) gives P(B)-P(A\cap B)=0, so P(A\cap B)=P(B).
Then P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(B)}{P(B)}=1 (provided P(B)\neq0).
That's the final chapter of the NCERT Class 12 Maths curriculum. Revisit any exercise for more practice, or explore solutions for the rest of the book.
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