Class 12 Maths NCERT Solutions Chapter 13 Ex 13.3 – Bayes' Theorem

Let R_1: first ball drawn is red, B_1: first ball drawn is black. P(R_1)=\dfrac{5}{10}=\dfrac12, P(B_1)=\dfrac12.

If the first ball was red, the urn now has 7 red and 5 black (12 total), so P(\text{2nd red}|R_1)=\dfrac{7}{12}. If the first ball was black, the urn now has 5 red and 7 black (12 total), so P(\text{2nd red}|B_1)=\dfrac{5}{12}.

By the theorem of total probability, P(\text{2nd red})=\dfrac12\times\dfrac{7}{12}+\dfrac12\times\dfrac{5}{12}=\dfrac{7}{24}+\dfrac{5}{24}=\dfrac{12}{24}=\dfrac12.

Let E_1: bag I is chosen, E_2: bag II is chosen, both with P(E_1)=P(E_2)=\dfrac12. Let A: the ball drawn is red.

P(A|E_1)=\dfrac48=\dfrac12 and P(A|E_2)=\dfrac28=\dfrac14.

By Bayes' theorem, P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}=\dfrac{\frac12\times\frac12}{\frac12\times\frac12+\frac12\times\frac14}=\dfrac{1/4}{1/4+1/8}=\dfrac{1/4}{3/8}=\dfrac23.

Let E_1: student is a hosteller, E_2: student is a day scholar. P(E_1)=0.6, P(E_2)=0.4.

Let A: student attains an A grade, with P(A|E_1)=0.3 and P(A|E_2)=0.2.

By Bayes' theorem, P(E_1|A)=\dfrac{0.6\times0.3}{0.6\times0.3+0.4\times0.2}=\dfrac{0.18}{0.18+0.08}=\dfrac{0.18}{0.26}=\dfrac{9}{13}.

Let E_1: student knows the answer, E_2: student guesses. P(E_1)=\dfrac34, P(E_2)=\dfrac14.

Let A: the answer is correct, with P(A|E_1)=1 (a student who knows answers correctly) and P(A|E_2)=\dfrac14.

By Bayes' theorem, P(E_1|A)=\dfrac{\frac34\times1}{\frac34\times1+\frac14\times\frac14}=\dfrac{3/4}{3/4+1/16}=\dfrac{3/4}{13/16}=\dfrac34\times\dfrac{16}{13}=\dfrac{12}{13}.

Let E: the person has the disease, so P(E)=0.001 and P(E')=0.999. Let A: the test is positive, with P(A|E)=0.99 and P(A|E')=0.005.

By Bayes' theorem, P(E|A)=\dfrac{0.001\times0.99}{0.001\times0.99+0.999\times0.005}=\dfrac{0.00099}{0.00099+0.004995}=\dfrac{0.00099}{0.005985}.

This simplifies to \dfrac{22}{133}\approx0.166 — even with a highly accurate test, the disease is so rare that a positive result still only makes actually having it about 16.6% likely.

Let E_1,E_2,E_3 be the events of choosing the two-headed, biased, and unbiased coin respectively, each with prior P(E_i)=\dfrac13.

Let A: the coin shows heads, with P(A|E_1)=1, P(A|E_2)=0.75, P(A|E_3)=0.5.

By Bayes' theorem, P(E_1|A)=\dfrac{\frac13\times1}{\frac13\times1+\frac13\times0.75+\frac13\times0.5}=\dfrac{1}{1+0.75+0.5}=\dfrac{1}{2.25}=\dfrac49.

Out of 2000+4000+6000=12000 insured drivers, P(\text{scooter})=\dfrac16, P(\text{car})=\dfrac13, P(\text{truck})=\dfrac12.

The accident probabilities given each type are 0.01, 0.03 and 0.15 respectively.

By Bayes' theorem, P(\text{scooter}|\text{accident})=\dfrac{\frac16\times0.01}{\frac16\times0.01+\frac13\times0.03+\frac12\times0.15}.

Using a common denominator of 600: the numerator is \dfrac{1}{600}, and the denominator sums to \dfrac{1}{600}+\dfrac{6}{600}+\dfrac{45}{600}=\dfrac{52}{600}. So the probability is \dfrac{1}{52}.

Let E_1: item made by A, E_2: item made by B. P(E_1)=0.6, P(E_2)=0.4. Let D: item is defective, with P(D|E_1)=0.02 and P(D|E_2)=0.01.

By Bayes' theorem, P(E_2|D)=\dfrac{0.4\times0.01}{0.6\times0.02+0.4\times0.01}=\dfrac{0.004}{0.012+0.004}=\dfrac{0.004}{0.016}=0.25.

Let E_1: first group wins, E_2: second group wins. P(E_1)=0.6, P(E_2)=0.4.

Let A: a new product is introduced, with P(A|E_1)=0.7 and P(A|E_2)=0.3.

By Bayes' theorem, P(E_2|A)=\dfrac{0.4\times0.3}{0.6\times0.7+0.4\times0.3}=\dfrac{0.12}{0.42+0.12}=\dfrac{0.12}{0.54}=\dfrac29.

Let E_1: die shows 5 or 6 (leading to three coin tosses), E_2: die shows 1, 2, 3 or 4 (leading to one coin toss). P(E_1)=\dfrac13, P(E_2)=\dfrac23.

Let A: exactly one head is obtained. In three tosses, P(A|E_1)=\binom31\left(\dfrac12\right)^3=\dfrac38. In one toss, "exactly one head" simply means the coin shows heads, so P(A|E_2)=\dfrac12.

By Bayes' theorem, P(E_2|A)=\dfrac{\frac23\times\frac12}{\frac13\times\frac38+\frac23\times\frac12}=\dfrac{1/3}{1/8+1/3}=\dfrac{1/3}{3/24+8/24}=\dfrac{1/3}{11/24}=\dfrac13\times\dfrac{24}{11}=\dfrac{8}{11}.

Let E_1,E_2,E_3 be the events "item produced by A, B, C" with priors P(E_1)=0.5, P(E_2)=0.3, P(E_3)=0.2.

Let D: item is defective, with P(D|E_1)=0.01, P(D|E_2)=0.05, P(D|E_3)=0.07.

P(D)=0.5\times0.01+0.3\times0.05+0.2\times0.07=0.005+0.015+0.014=0.034.

By Bayes' theorem, P(E_1|D)=\dfrac{0.005}{0.034}=\dfrac{5}{34}.

Let D: the lost card is a diamond, so P(D)=\dfrac{13}{52}=\dfrac14 and P(D')=\dfrac34. Let A: both cards drawn from the remaining 51 are diamonds.

If the lost card was a diamond, 12 diamonds remain among 51 cards, so P(A|D)=\dfrac{12}{51}\times\dfrac{11}{50}=\dfrac{132}{2550}. If not, 13 diamonds remain among 51, so P(A|D')=\dfrac{13}{51}\times\dfrac{12}{50}=\dfrac{156}{2550}.

By Bayes' theorem, P(D|A)=\dfrac{\frac14\times\frac{132}{2550}}{\frac14\times\frac{132}{2550}+\frac34\times\frac{156}{2550}}=\dfrac{132}{132+3\times156}=\dfrac{132}{132+468}=\dfrac{132}{600}=\dfrac{11}{50}.

Let H: a head actually occurred, so P(H)=P(H')=\dfrac12. Let R: A reports a head.

If a head occurred, A truthfully reports it with probability \dfrac45; if a tail occurred, A would report a head only by lying, with probability \dfrac15.

By Bayes' theorem, P(H|R)=\dfrac{\frac12\times\frac45}{\frac12\times\frac45+\frac12\times\frac15}=\dfrac{4/10}{4/10+1/10}=\dfrac{4}{5}.

Since A\subset B, every outcome of A is also in B, so A\cap B=A, giving P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(A)}{P(B)}.

Since 0 \lt P(B)\le1, dividing P(A) by P(B) can only keep it the same (if P(B)=1) or make it larger (if P(B) \lt 1). So P(A|B)\ge P(A) always.